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Animated Solution for Mathematics - Conic Sections: The locus of the foot of perpendicular drawn from the centre of the ellipse $x^{2} + 3 y^{2} = 6$ on any tangent to it is

Q: What happens to the $\pm$ sign when we square the equation?

It vanishes! When you square both sides, the term $(\pm \sqrt{...})^2$ becomes just the expression inside the square root. This is a beautiful moment in algebra where the ambiguity of the 'plus or minus' is resolved by the squaring process, leaving us with a clean, definitive equation for the locus.

Select Answer:

Visualized Solution

Standard Form of the Ellipse

Given equation: $x^{2} + 3 y^{2} = 6$
Divide by $6$ to get standard form: $\frac{x ^{2}}{6} + \frac{y ^{2}}{2} = 1$

Ellipse Parameters

Comparing with $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ :
$a^{2} = 6$
$b^{2} = 2$

Equation of Tangent

General equation of tangent in slope form:
$y = m x \pm a^{2} m^{2} + b^{2}$

Tangent for Given Ellipse

Substitute $a^{2} = 6$ and $b^{2} = 2$ :
$y = m x \pm 6 m^{2} + 2$

Foot of Perpendicular

Let $P (h, k)$ be the foot of the perpendicular.
Drawn from the center $(0, 0)$ to the tangent.

Slope Relationship

The line $O P$ passes through $(0, 0)$ and $P (h, k)$ .
Slope of $O P = \frac{k - 0}{h - 0} = \frac{k}{h}$

Slope of Tangent $m$

$O P$ is perpendicular to the tangent.
Product of slopes $= - 1$
Slope of tangent $m = - \frac{h}{k}$

Substituting $m$ and Point $P$

Point $P (h, k)$ lies on the tangent.
Substitute $x = h, y = k$ and $m = - \frac{h}{k}$ into:
$y = m x \pm 6 m^{2} + 2$
$k = (- \frac{h}{k}) h \pm 6 (- \frac{h}{k})^{2} + 2$

Rearranging the Equation

Simplify the terms:
$k = - \frac{h ^{2}}{k} \pm \frac{6 h ^{2}}{k ^{2}} + 2$
Move $- \frac{h ^{2}}{k}$ to the left side:
$k + \frac{h ^{2}}{k} = \pm \frac{6 h ^{2}}{k ^{2}} + 2$

Simplifying the Terms

Take common denominator $k$ on both sides:
$\frac{k ^{2} + h ^{2}}{k} = \pm \frac{6 h ^{2} + 2 k ^{2}}{k ^{2}}$

Squaring Both Sides

Square both sides to remove the square root:
$(\frac{k ^{2} + h ^{2}}{k})^{2} = \frac{6 h ^{2} + 2 k ^{2}}{k ^{2}}$
$\frac{( k ^{2} + h ^{2} ) ^{2}}{k ^{2}} = \frac{6 h ^{2} + 2 k ^{2}}{k ^{2}}$

Canceling Denominators

Cancel $k^{2}$ from both sides:
$(h^{2} + k^{2})^{2} = 6 h^{2} + 2 k^{2}$

Final Locus Equation

Replace $(h, k)$ with $(x, y)$ to get the locus:
$(x^{2} + y^{2})^{2} = 6 x^{2} + 2 y^{2}$
Correct Option: (a)

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

The given equation of the ellipse is

x^{2} + 3 y^{2} = 6

. To reveal its standard form, we divide the entire equation by

6

\frac{x ^{2}}{6} + \frac{y ^{2}}{2} = 1

From this, we identify the parameters

a^{2} = 6

and

b^{2} = 2

. These values define the geometry of our ellipse.

The Tangent Equation

Any tangent line to this ellipse can be represented using the slope-intercept form:

y = m x \pm a^{2} m^{2} + b^{2}

Substituting our known parameters, the equation of the tangent becomes:

y = m x \pm 6 m^{2} + 2

The Perpendicular Condition

We are interested in the foot of the perpendicular

P (h, k)

dropped from the origin

(0, 0)

to this tangent line. The slope of the line segment

O P

is given by

\frac{k}{h}

Since

O P

is perpendicular to the tangent, the slope of the tangent

m

must be the negative reciprocal of the slope of

O P

m = - \frac{h}{k}

Deriving the Locus

The point

P (h, k)

must lie on the tangent line. Substituting

m = - \frac{h}{k}

into the tangent equation, we obtain:

k = (- \frac{h}{k}) h \pm 6 (- \frac{h}{k})^{2} + 2

Rearranging the terms to isolate the radical, we get:

k + \frac{h ^{2}}{k} = \pm \frac{6 h ^{2}}{k ^{2}} + 2

Combining the terms on the left side over a common denominator

k

yields:

\frac{k ^{2} + h ^{2}}{k} = \pm \frac{6 h ^{2} + 2 k ^{2}}{k ^{2}}

Final Calculation

To eliminate the radical, we square both sides of the equation:

\frac{( h ^{2} + k ^{2} ) ^{2}}{k ^{2}} = \frac{6 h ^{2} + 2 k ^{2}}{k ^{2}}

Multiplying both sides by

k^{2}

and replacing the coordinates

(h, k)

with the general variables

(x, y)

, we arrive at the final locus:

(x^{2} + y^{2})^{2} = 6 x^{2} + 2 y^{2}

This equation represents the pedal curve of the ellipse with respect to its center.

Common Traps & FAQs

Why do we use the slope form of the tangent instead of the point form?

Great question! The point form requires us to know the point of tangency, which we don't have. The slope form is a 'universal' tool—it allows us to define any tangent based solely on its slope

m

. Since we are relating the slope of the perpendicular to the tangent, the slope form is the most direct path to the solution.

What happens to the

\pm

sign when we square the equation?

It vanishes! When you square both sides, the term

(\pm ...)^{2}

becomes just the expression inside the square root. This is a beautiful moment in algebra where the ambiguity of the 'plus or minus' is resolved by the squaring process, leaving us with a clean, definitive equation for the locus.

Is there a geometric name for this locus?

Yes! This locus is known as the 'Pedal Curve' of the ellipse with respect to its center. Specifically, for an ellipse, this curve is a special case of a quartic curve. It is fascinating to see how a simple geometric constraint leads to such a structured algebraic result.

Animated Solution for Mathematics - Conic Sections: The locus of the foot of perpendicular drawn from the centre of the ellipse $x^{2} + 3 y^{2} = 6$ on any tangent to it is

Select Answer:

Visualized Solution

Standard Form of the Ellipse

Ellipse Parameters

Equation of Tangent

Tangent for Given Ellipse

Foot of Perpendicular

Slope Relationship

Slope of Tangent $m$

Substituting $m$ and Point $P$

Rearranging the Equation

Simplifying the Terms

Squaring Both Sides

Canceling Denominators

Final Locus Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Equation

The Perpendicular Condition

Deriving the Locus

Final Calculation

Similar Questions

Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ from any of its foci?

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

Comprehension Passage

The coordinates of $A$ and $B$ are

The orthocenter of the triangle $P A B$ is

The equation of the locus of the point whose distances from the point $P$ and the line $A B$ are equal, is

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ for some $a \in R$ , then the distance between the foci of the ellipse is:

The angle between a pair of tangents drawn from a point $P$ to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 sin^{2} α + 13 cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ , for some $a \in R$ then the distance between the foci of the ellipse is :

A tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the ellipse $x^{2} + 2 y^{2} = 6$ at $P$ and $Q$ . Prove that the tangents at $P$ and $Q$ of the ellipse $x^{2} + 2 y^{2} = 6$ are at right angles.

If a line, $y = m x + c$ is a tangent to the circle, $(x - 3)^{2} + y^{2} = 1$ and it is perpendicular to a line $L_{1}$ , where $L_{1}$ is the tangent to the circle, $x^{2} + y^{2} = 1$ at the point $(\frac{1}{2}, \frac{1}{2})$ ; then:

Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6 on any tangent to it is

Select Answer:

Visualized Solution

Standard Form of the Ellipse

Ellipse Parameters

Equation of Tangent

Tangent for Given Ellipse

Foot of Perpendicular

Slope Relationship

Slope of Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Rearranging the Equation

Simplifying the Terms

Squaring Both Sides

Canceling Denominators

Final Locus Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Equation

The Perpendicular Condition

Deriving the Locus

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, 4x2​+2y2​=1 from any of its foci?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If tangents are drawn to the ellipse x2+2y2=2, then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coordinates of A and B are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The orthocenter of the triangle PAB is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the locus of the point whose distances from the point P and the line AB are equal, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 3x+4y=122​ is a tangent to the ellipse a2x2​+9y2​=1 for some a∈R, then the distance between the foci of the ellipse is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between a pair of tangents drawn from a point P to the circle x2+y2+4x−6y+9sin2α+13cos2α=0 is 2α. The equation of the locus of the point P is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 3x+4y=122​ is a tangent to the ellipse a2x2​+9y2​=1, for some a∈R then the distance between the foci of the ellipse is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A tangent to the ellipse x2+4y2=4 meets the ellipse x2+2y2=6 at P and Q. Prove that the tangents at P and Q of the ellipse x2+2y2=6 are at right angles.

Animated Solution for Mathematics - Conic Sections: The locus of the foot of perpendicular drawn from the centre of the ellipse $x^{2} + 3 y^{2} = 6$ on any tangent to it is

Slope of Tangent $m$

Substituting $m$ and Point $P$

Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ from any of its foci?

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

The coordinates of $A$ and $B$ are

The orthocenter of the triangle $P A B$ is

The equation of the locus of the point whose distances from the point $P$ and the line $A B$ are equal, is

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ for some $a \in R$ , then the distance between the foci of the ellipse is:

The angle between a pair of tangents drawn from a point $P$ to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 sin^{2} α + 13 cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

If $3 x + 4 y = 122$ is a tangent to the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{9} = 1$ , for some $a \in R$ then the distance between the foci of the ellipse is :

A tangent to the ellipse $x^{2} + 4 y^{2} = 4$ meets the ellipse $x^{2} + 2 y^{2} = 6$ at $P$ and $Q$ . Prove that the tangents at $P$ and $Q$ of the ellipse $x^{2} + 2 y^{2} = 6$ are at right angles.