Two tangent lines l1 and l2 are drawn from the point (2, 0) to the parabola 2y² = -x. If the lines l

Q: Wait, why did we use $c = \frac{a}{m}$ instead of solving the quadratic equation formed by substituting the line into the parabola?

That is a fantastic question! You absolutely *could* substitute the line equation into the parabola and set the discriminant to zero. However, using the standard condition $c = \frac{a}{m}$ is a powerful shortcut that saves time and reduces the chance of algebraic errors. In JEE Advanced, speed and accuracy are your best friends!

Q: Why is the radius $\sqrt{r}$ and not $r$? I always get confused by the circle equation.

It is a common trap! Always look closely at the RHS of the circle equation. If the equation is $(x - h)^2 + (y - k)^2 = R^2$, then the radius is $R$. Here, the equation is $(x - 5)^2 + y^2 = r$, so the radius squared is $r$, which means the radius itself is $\sqrt{r}$. Always verify the form before proceeding.

Q: Does it matter which tangent line I choose to calculate the radius?

In this specific problem, due to the symmetry of the parabola and the circle's center lying on the axis of symmetry, both tangents will yield the same distance. You can pick either $x - 4y - 2 = 0$ or $x + 4y - 2 = 0$. The result will be identical!

Question

Two tangent lines $l_1$ and $l_2$ are drawn from the point $(2, 0)$ to the parabola $2y^2 = -x$. If the lines $l_1$ and $l_2$ are also tangent to the circle $(x - 5)^2 + y^2 = r$, then $17r$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to unravel a beautiful problem that bridges the gap between the parabolic world and the circular realm.

Imagine you are standing on the Cartesian plane. You have a parabola, $2y^2 = -x$, opening its arms towards the negative $x$-axis. You are holding a point $P(2, 0)$ and drawing two lines that kiss this parabola perfectly.

These lines, $l_1$ and $l_2$, are not just tangents; they are the keys to unlocking the radius of a circle centered at $(5, 0)$. Let's embark on this journey.

### Standardizing the Parabola

Before we can use our powerful toolkit of conic section formulas, we must bring our parabola into its standard form. We are given $2y^2 = -x$.

Dividing by $2$, we get:

$$y^2 = -\frac{1}{2}x$$

Comparing this to the standard equation $y^2 = 4ax$, we immediately see that $4a = -\frac{1}{2}$, which gives us $a = -\frac{1}{8}$. This negative value of $a$ confirms our intuition: the parabola opens to the left.

### The Slope Game

Now, we need the equations of those tangent lines. For any parabola $y^2 = 4ax$, the equation of a tangent with slope $m$ is given by:

$$y = mx + \frac{a}{m}$$

Substituting our value of $a$, we get $y = mx - \frac{1}{8m}$. This equation represents the entire family of tangents.

Since these lines pass through $P(2, 0)$, we substitute $x = 2$ and $y = 0$ into our tangent equation:

$$0 = 2m - \frac{1}{8m}$$

Solving for $m$, we find $2m = \frac{1}{8m}$, which leads to $16m^2 = 1$, or $m = \pm \frac{1}{4}$.

The two lines are $y = \frac{1}{4}x - \frac{1}{2}$ and $y = -\frac{1}{4}x + \frac{1}{2}$, which simplify to:

$$x - 4y - 2 = 0 \quad 	ext{and} \quad x + 4y - 2 = 0$$

### The Circle and the Distance Bridge

Finally, we turn to the circle $(x - 5)^2 + y^2 = r$. Its center is $C(5, 0)$ and its radius is $\sqrt{r}$.

The problem states that our tangent lines are also tangent to this circle. This means the perpendicular distance from the center $C(5, 0)$ to the line $x - 4y - 2 = 0$ must equal the radius $\sqrt{r}$.

Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$, we calculate:

$$\sqrt{r} = \frac{|1(5) - 4(0) - 2|}{\sqrt{1^2 + (-4)^2}}$$

This simplifies to:

$$\sqrt{r} = \frac{3}{\sqrt{17}}$$

Squaring both sides, we find $r = \frac{9}{17}$.

### Final Calculation

The question asks for the value of $17r$. Multiplying our result by $17$:

$$17 	imes \frac{9}{17} = 9$$

The elegance of this cancellation is the hallmark of a well-crafted JEE problem. You have successfully navigated the geometry and arrived at the final answer of **9**.

Animated Solution for Mathematics - Conic Sections: Two tangent lines $l_{1}$ and $l_{2}$ are drawn from the point $(2, 0)$ to the parabola $2 y^{2} = - x$ . If the lines $l_{1}$ and $l_{2}$ are also tangent to the circle $(x - 5)^{2} + y^{2} = r$ , then $17 r$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

Standard Form of Parabola

General Tangent Equation

Applying Point $P (2, 0)$

Solving for Slope $m$

Equations of Tangent Lines

Circle Center and Radius

Tangency Condition for Circle

Substituting into Distance Formula

Calculating $r$

Final Result: $17 r$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Standardizing the Parabola

The Slope Game

The Circle and the Distance Bridge

Final Calculation

Similar Questions

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A triangle is formed by the tangents at the point $(2, 2)$ on the curves $y^{2} = 2 x$ and $x^{2} + y^{2} = 4 x$ , and the line $x + y + 2 = 0$ . If r is the radius of its circumcircle, then $r^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

Standard Form of Parabola

General Tangent Equation

Applying Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2,0)

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

Equations of Tangent Lines

Circle Center and Radius

Tangency Condition for Circle

Substituting into Distance Formula

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

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 17r

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Standardizing the Parabola

The Slope Game

The Circle and the Distance Bridge

Final Calculation

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