Let a line perpendicular to the line 2x - y = 10 touch the parabola y² = 4(x - 9) at the point P. Th

Q: Why do we need to shift the parabola? Can't we just differentiate the original equation?

You absolutely can! Differentiation is a powerful tool. However, in JEE Advanced, time is currency. Using the standard form transformation $X = x - 9$ allows you to use the standard point-of-contact formula $(\frac{a}{m^2}, \frac{2a}{m})$ directly, which is much faster and less prone to algebraic errors than implicit differentiation.

Q: I got the slope as $2$. Why is the tangent slope $-1/2$?

Great catch! The line $2x - y = 10$ has a slope of $2$. Since the tangent is *perpendicular* to this line, we use the condition $m_1 \cdot m_2 = -1$. Thus, $2 \cdot m_2 = -1$, leading to $m_2 = -\frac{1}{2}$. Always visualize the perpendicularity!

Q: How do I know if the point $P$ is correct?

Once you find $P(13, -4)$, you can verify it by checking if it satisfies the parabola equation $y^2 = 4(x - 9)$. $(-4)^2 = 16$ and $4(13 - 9) = 4(4) = 16$. Since $16 = 16$, your point is definitely on the parabola. Trust your math!

Question

Let a line perpendicular to the line $2x - y = 10$ touch the parabola $y^2 = 4(x - 9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2 + y^2 - 14x - 8y + 56 = 0$ is __________

Sigma Percentile · Accepted Answer

### The Geometry of Precision: A Journey to Point $P$

Welcome, future engineer. Today, we are not just solving a problem; we are orchestrating a symphony of coordinate geometry. We have a parabola, a line, and a circle.

Our mission is to find a specific point $P$ on the parabola where the tangent is perfectly perpendicular to a given line, and then calculate its distance to the center of a circle. This is a classic JEE Advanced challenge that tests your ability to synthesize multiple concepts. Let us begin.

### Phase 1: The Tangent's Identity

First, we must understand the line we are dealing with. The equation is $2x - y = 10$.

To find its slope, we rearrange it into the slope-intercept form, $y = mx + c$. This gives us $y = 2x - 10$. Clearly, the slope $m_1$ is $2$.

Now, the problem states that our tangent is perpendicular to this line. Recall the fundamental condition for perpendicularity: the product of the slopes must be $-1$.

Therefore, if $m_1 = 2$, our tangent slope $m$ must satisfy $2 \cdot m = -1$, which gives us:

$$m = -\frac{1}{2}$$

This is our first victory.

### Phase 2: Taming the Parabola

The parabola given is $y^2 = 4(x - 9)$. It is not centered at the origin, which might look intimidating.

But fear not! We use a simple transformation. Let $X = x - 9$ and $Y = y$.

Now, the equation becomes $Y^2 = 4X$. This is the standard form $Y^2 = 4aX$, where $4a = 4$, implying $a = 1$. By shifting our perspective, we have simplified the problem significantly.

### Phase 3: The Point of Contact

With the parabola in standard form and the slope $m = -\frac{1}{2}$ in hand, we can use the elegant point-of-contact formula. For a parabola $Y^2 = 4aX$, the point of contact for a tangent with slope $m$ is:

$$(X, Y) = \left( \frac{a}{m^2}, \frac{2a}{m} \right)$$

Substituting our values, $a = 1$ and $m = -\frac{1}{2}$, we get:

$$X = \frac{1}{(-\frac{1}{2})^2} = \frac{1}{\frac{1}{4}} = 4$$

$$Y = \frac{2(1)}{-\frac{1}{2}} = -4$$

We have found the coordinates in our transformed system! Now, we map them back: $x - 9 = 4 \Rightarrow x = 13$, and $y = -4$. Thus, our point $P$ is $(13, -4)$.

### Phase 4: The Circle's Heart

Next, we look at the circle: $x^2 + y^2 - 14x - 8y + 56 = 0$. To find the center, we compare this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

The center is $(-g, -f)$. Here, $2g = -14 \Rightarrow g = -7$, and $2f = -8 \Rightarrow f = -4$.

Thus, the center $C$ is $(7, 4)$.

### Phase 5: The Final Distance

We are at the finish line. We need the distance between $P(13, -4)$ and $C(7, 4)$.

Using the distance formula $CP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, we calculate:

$$CP = \sqrt{(13 - 7)^2 + (-4 - 4)^2}$$

$$CP = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

The distance is exactly **10**. You have successfully navigated the complexity of this problem. Keep this clarity of thought, and you will conquer any challenge the JEE throws your way.

Animated Solution for Mathematics - Conic Sections: Let a line perpendicular to the line $2 x - y = 10$ touch the parabola $y^{2} = 4 (x - 9)$ at the point $P$ . The distance of the point $P$ from the centre of the circle $x^{2} + y^{2} - 14 x - 8 y + 56 = 0$ is __________

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

Slope of the Given Line

Slope of the Perpendicular Tangent

Parabola Transformation

Point of Contact Formula

Calculating Transformed Coordinates

Finding Point $P$

Center of the Circle

The Distance Formula

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Precision

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

Let the shortest distance from $(a, 0)$ , $a > 0$ to the parabola $y^{2} = 4 x$ be 4. Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is:

Let P be a point on the parabola, $x^{2} = 4 y$ . If the distance of P from the centre of the circle, $x^{2} + y^{2} + 6 x + 8 = 0$ is minimum, then the equation of the tangent to the parabola at P, is :

Let $P$ be the point on the parabola $y^{2} = 4 x$ which is at the shortest distance from the center $S$ of the circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0$ . Let $Q$ be the point on the circle dividing the line segment $S P$ internally. Then

Let the tangent to the parabola $y^{2} = 12 x$ at the point $(3, α)$ be perpendicular to the line $2 x + 2 y = 3$ . Then the square of distance of the point $(6, - 4)$ from the normal to the hyperbola $α^{2} x^{2} - 9 y^{2} = 9 α^{2}$ at its point $(α - 1, α + 2)$ is equal to .............

A circle touches the line $y = x$ at a point $P$ such that $O P = 42$ , where $O$ is the origin. The circle contains the point $(- 10, 2)$ in its interior and the length of its chord on the line $x + y = 0$ is $62$ . Determine the equation of the circle.

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

Let $P$ be the point on the parabola, $y^{2} = 8 x$ which is at a minimum distance from the centre $C$ of the circle, $x^{2} + (y + 6)^{2} = 1$ . Then the equation of the circle, passing through $C$ and having its centre at $P$ is:

Let $A B$ be a chord of length 12 of the circle $(x - 2)^{2} + (y + 1)^{2} = \frac{169}{4}$ . If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$ , then five times the distance of point $P$ from chord $A B$ is equal to ________

Find the coordinates of the point at which the circles $x^{2} + y^{2} - 4 x - 2 y - 4 = 0$ and $x^{2} + y^{2} - 12 x - 8 y - 36 = 0$ touch each other. Also find equations common tangents touching the circles in the distinct points.