JEE Main 2019 (10 April Shift 2)

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Animated Solution for Mathematics - Conic Sections: If $5 x + 9 = 0$ is the directrix of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is :

Q: Wait, why do we reject the positive root for the directrix?

We do not reject it! The hyperbola has two directrices: $x = \frac{a}{e}$ and $x = -\frac{a}{e}$. The problem gives us $5x + 9 = 0$, which simplifies to $x = -\frac{9}{5}$. Since this is the negative directrix, we must choose the corresponding negative focus.

Q: How do I know which focus corresponds to which directrix?

Great question. In a standard hyperbola, the directrix $x = \frac{a}{e}$ is associated with the focus $(ae, 0)$, and the directrix $x = -\frac{a}{e}$ is associated with the focus $(-ae, 0)$. They always stay on the same side of the center.

Select Answer:

Visualized Solution

The Hyperbola Equation

Given equation: $16 x^{2} - 9 y^{2} = 144$
We need to find the focus corresponding to the directrix $5 x + 9 = 0$ .

Standard Form Conversion

Divide both sides by $144$ : $\frac{16 x ^{2}}{144} - \frac{9 y ^{2}}{144} = 1$
Simplify fractions: $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$
This is the standard form of a hyperbola.

Identifying Parameters $a$ and $b$

Compare with standard equation: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$
Value of a: $a^{2} = 9 ⟹ a = 3$
Value of b: $b^{2} = 16 ⟹ b = 4$

Eccentricity Formula

Eccentricity formula for hyperbola: $e = 1 + \frac{b ^{2}}{a ^{2}}$
This determines the shape and spread of the hyperbola.

Calculating Eccentricity $e$

Substitute values: $e = 1 + \frac{16}{9}$
Take LCM: $e = \frac{25}{9}$
Eccentricity: $e = \frac{5}{3}$

Coordinates of Foci

Foci formula: $(\pm a e, 0)$
Substitute $a = 3$ and $e = \frac{5}{3}$
Calculate: $(\pm 3 \cdot \frac{5}{3}, 0)$
Foci Coordinates: $(\pm 5, 0)$

Equations of Directrices

Directrices formula: $x = \pm \frac{a}{e}$
Substitute $a = 3$ and $e = \frac{5}{3}$
Calculate: $x = \pm \frac{3}{5/3}$
Directrices Equations: $x = \pm \frac{9}{5}$

Analyzing the Given Directrix

Given directrix equation: $5 x + 9 = 0$
Rearrange terms: $5 x = - 9$
Specific Directrix: $x = - \frac{9}{5}$

Matching Focus to Directrix

The directrix $x = - \frac{a}{e}$ corresponds to the focus $(- a e, 0)$ .
Therefore, $x = - \frac{9}{5}$ corresponds to the focus $(- 5, 0)$ .
Final Answer: $(- 5, 0)$

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The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Setup

To understand the geometry of the hyperbola defined by

16 x^{2} - 9 y^{2} = 144

, we must first convert the equation into its standard form. By dividing the entire equation by

144

, we obtain:

\frac{16 x ^{2}}{144} - \frac{9 y ^{2}}{144} = 1

This simplifies to the standard form:

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

Comparing this to the general form

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

, we identify the parameters

a^{2} = 9

and

b^{2} = 16

, which yields

a = 3

and

b = 4

The Eccentricity

The Measure of Stretch

The eccentricity

e

defines the shape of the hyperbola. It is calculated using the formula:

e = 1 + \frac{b ^{2}}{a ^{2}}

Substituting our known values into this expression, we find:

e = 1 + \frac{16}{9} = \frac{25}{9} = \frac{5}{3}

This value,

e = \frac{5}{3}

, is essential for determining the coordinates of the foci, which are located at

(\pm a e, 0)

. Calculating these, we get

(\pm 3 \cdot \frac{5}{3}, 0)

, resulting in the foci at

(\pm 5, 0)

The Invisible Anchors

Directrices

The directrices of the hyperbola are defined by the vertical lines

x = \pm \frac{a}{e}

. Using our values for

a

and

e

x = \pm \frac{3}{5/3} = \pm \frac{9}{5}

The problem specifies the directrix

5 x + 9 = 0

, which rearranges to

x = - \frac{9}{5}

. This corresponds to the left-hand directrix.

Final Calculation

In a hyperbola, the directrix and its corresponding focus must lie on the same side of the center. Since the directrix

x = - \frac{9}{5}

is on the negative side of the

y

-axis, we select the focus with the negative

x

-coordinate.

Therefore, the focus corresponding to the directrix

5 x + 9 = 0

is $(- 5, 0)$ .

Common Traps & FAQs

Wait, why do we reject the positive root for the directrix?

We do not reject it! The hyperbola has two directrices:

x = \frac{a}{e}

and

x = - \frac{a}{e}

. The problem gives us

5 x + 9 = 0

, which simplifies to

x = - \frac{9}{5}

. Since this is the negative directrix, we must choose the corresponding negative focus.

How do I know which focus corresponds to which directrix?

Great question. In a standard hyperbola, the directrix

x = \frac{a}{e}

is associated with the focus

(a e, 0)

, and the directrix

x = - \frac{a}{e}

is associated with the focus

(- a e, 0)

. They always stay on the same side of the center.

Can I use the definition of a hyperbola to solve this?

Yes, you absolutely can! The definition states that the distance from a point on the hyperbola to the focus is

e

times the distance to the directrix. While that is a powerful tool, using the standard formulas for

a

e

, and the directrix is much faster for this specific problem.

Animated Solution for Mathematics - Conic Sections: If $5 x + 9 = 0$ is the directrix of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is :

Select Answer:

Visualized Solution

The Hyperbola Equation

Standard Form Conversion

Identifying Parameters $a$ and $b$

Eccentricity Formula

Calculating Eccentricity $e$

Coordinates of Foci

Equations of Directrices

Analyzing the Given Directrix

Matching Focus to Directrix

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Eccentricity

The Invisible Anchors

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 5x+9=0 is the directrix of the hyperbola 16x2−9y2=144, then its corresponding focus is :

Select Answer:

Visualized Solution

The Hyperbola Equation

Standard Form Conversion

Identifying Parameters .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Eccentricity Formula

Calculating Eccentricity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } e

Coordinates of Foci

Equations of Directrices

Analyzing the Given Directrix

Matching Focus to Directrix

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Eccentricity

The Invisible Anchors

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Conic Sections: If $5 x + 9 = 0$ is the directrix of the hyperbola $16 x^{2} - 9 y^{2} = 144$ , then its corresponding focus is :

Identifying Parameters $a$ and $b$

Calculating Eccentricity $e$

If the line $x - 1 = 0$ , is a directrix of the hyperbola $k x^{2} - y^{2} = 6$ , then the hyperbola passes through the point

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola $y^{2} = 4 a x$ is another parabola with directrix

The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16 x^{2} - 9 y^{2} + 32 x + 36 y - 164 = 0$ , and its foci is :

If a directrix of a hyperbola centred at the origin and passing through the point $(4, - 23)$ is $5 x = 45$ and its eccentricity is $e$ , then :

A parabola has the origin as its focus and the line $x = 2$ as the directrix. Then the vertex of the parabola is at

Let $P_{1}$ be a parabola with vertex $(3, 2)$ and focus $(4, 4)$ and $P_{2}$ be its mirror image with respect to the line $x + 2 y = 6$ . Then the directrix of $P_{2}$ is $x + 2y = ____$ .

The locus of the mid-point of the line segment joining the focus of the parabola $y^{2} = 4 a x$ to a moving point of the parabola, is another parabola whose directrix is: