JEE MainMathematics

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Length of a Normal Segment to a Hyperbola Intersecting the y-Axis

Q: What is the standard parametric equation of the normal to (x^2)/(a^2) - (y^2)/(b^2) = 1?

The normal at (asecθ, btanθ) is ax/secθ + by/tanθ = a^2 + b^2, which simplifies to axcosθ + bycotθ = a^2 + b^2.

The vertices of a hyperbola H are (± 6,0) and its eccentricity is (√(5))/2. Let N be the normal to H at a point in the first quadrant and parallel to the line √(2) x+y=2 √(2). If d is the length of the line segment of N between H and the y -axis then d^2 is equal to

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Step-by-step solutionView

Correct answer

The value of d² is 216.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Vertices of hyperbola $H$ are $(\pm 6, 0)$ , eccentricity $e = \frac{\sqrt{5}}{2}$ , normal $N$ is at a point $P$ in the first quadrant and parallel to $\sqrt{2} x + y = 2 \sqrt{2}$ . $d$ is the distance along $N$ from $P$ to the $y$ -axis.

02Goal

Find the value of $d^{2}$ , where $d$ is the distance between point $P$ on $H$ and the intersection of normal $N$ with the $y$ -axis.

03Approach

Find $a$ and $b$ to write the standard equation of $H$ . Write the equation of the normal in parametric form $(a \sec θ, b \tan θ)$ , equate its slope to $- \sqrt{2}$ to find $θ$ , then obtain the coordinates of $P$ and the $y$ -intercept of the normal, and compute $d^{2}$ .

04Execute

From vertices $(\pm a, 0) = (\pm 6, 0)$ , we have $a = 6$ . Using $b^{2} = a^{2} (e^{2} - 1)$ , we get $b^{2} = 36 (\frac{5}{4} - 1) = 36 \times \frac{1}{4} = 9$ , so $b = 3$ . The equation of $H$ is $\frac{x^{2}}{36} - \frac{y^{2}}{9} = 1$ .

05Execute

The equation of the normal at $P (a \sec θ, b \tan θ) = (6 \sec θ, 3 \tan θ)$ is $\frac{a x}{\sec θ} + \frac{b y}{\tan θ} = a^{2} + b^{2}$ , which gives $6 x \cos θ + 3 y \cot θ = 36 + 9 = 45$ .

06Execute

The slope of the normal is $m = - \frac{6 \cos θ}{3 \cot θ} = - 2 \sin θ$ . Since $N$ is parallel to $\sqrt{2} x + y = 2 \sqrt{2}$ , whose slope is $- \sqrt{2}$ , we have $- 2 \sin θ = - \sqrt{2} ⟹ \sin θ = \frac{1}{\sqrt{2}}$ . Since $P$ is in the first quadrant, $θ = \frac{π}{4}$ .

07Execute

For $θ = \frac{π}{4}$ , the point of contact is $P (6 \sec \frac{π}{4}, 3 \tan \frac{π}{4}) = (6 \sqrt{2}, 3)$ . The equation of the normal becomes $6 x (\frac{1}{\sqrt{2}}) + 3 y (1) = 45 ⟹ 3 \sqrt{2} x + 3 y = 45 ⟹ \sqrt{2} x + y = 15$ . The $y$ -axis intercept $K$ is $(0, 15)$ .

08Execute

The distance squared between $P (6 \sqrt{2}, 3)$ and $K (0, 15)$ is $d^{2} = (6 \sqrt{2} - 0)^{2} + (3 - 15)^{2} = 72 + (- 12)^{2} = 72 + 144 = 216$ .

✓Verify

Check that $P (6 \sqrt{2}, 3)$ lies on $H$ : $\frac{(6 \sqrt{2})^{2}}{36} - \frac{3^{2}}{9} = \frac{72}{36} - \frac{9}{9} = 2 - 1 = 1$ , which is correct. Normal at $(x_{1}, y_{1})$ has slope $- \frac{a^{2} y_{1}}{b^{2} x_{1}} = - \frac{36 \times 3}{9 \times 6 \sqrt{2}} = - \frac{108}{54 \sqrt{2}} = - \sqrt{2}$ , matching the given slope.

Hints that build this answer step by step

Given the vertices (±6, 0) and eccentricity e = √5 / 2, what is the standard equation of the hyperbola H?
x²/36 - y²/9 = 1
Using the parametric point (6 sec θ, 3 tan θ) on H, what is the value of θ if the normal has slope -√2?
θ = π/4
What are the coordinates of the point of contact P on H and the intersection Q of the normal with the y-axis?
P = (6√2, 3) and Q = (0, 15)
What is the square of the distance d between P(6√2, 3) and Q(0, 15)?
d² = 216

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

What is the standard parametric equation of the normal to

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

The normal at $(a \sec θ, b \tan θ)$ is $\frac{a x}{\sec θ} + \frac{b y}{\tan θ} = a^{2} + b^{2}$ , which simplifies to $a x \cos θ + b y \cot θ = a^{2} + b^{2}$ .

Answer

Accepted answer: 216

The value of d² is 216.

Why each option works or fails

Step-by-step solution

Given: Vertices of hyperbola $H$ are $(\pm 6, 0)$ , eccentricity $e = \frac{\sqrt{5}}{2}$ , normal $N$ is at a point $P$ in the first quadrant and parallel to $\sqrt{2} x + y = 2 \sqrt{2}$ . $d$ is the distance along $N$ from $P$ to the $y$ -axis.
Goal: Find the value of $d^{2}$ , where $d$ is the distance between point $P$ on $H$ and the intersection of normal $N$ with the $y$ -axis.
Approach: Find $a$ and $b$ to write the standard equation of $H$ . Write the equation of the normal in parametric form $(a \sec θ, b \tan θ)$ , equate its slope to $- \sqrt{2}$ to find $θ$ , then obtain the coordinates of $P$ and the $y$ -intercept of the normal, and compute $d^{2}$ .
Execute: From vertices $(\pm a, 0) = (\pm 6, 0)$ , we have $a = 6$ . Using $b^{2} = a^{2} (e^{2} - 1)$ , we get $b^{2} = 36 (\frac{5}{4} - 1) = 36 \times \frac{1}{4} = 9$ , so $b = 3$ . The equation of $H$ is $\frac{x^{2}}{36} - \frac{y^{2}}{9} = 1$ .
Execute: The equation of the normal at $P (a \sec θ, b \tan θ) = (6 \sec θ, 3 \tan θ)$ is $\frac{a x}{\sec θ} + \frac{b y}{\tan θ} = a^{2} + b^{2}$ , which gives $6 x \cos θ + 3 y \cot θ = 36 + 9 = 45$ .
Execute: The slope of the normal is $m = - \frac{6 \cos θ}{3 \cot θ} = - 2 \sin θ$ . Since $N$ is parallel to $\sqrt{2} x + y = 2 \sqrt{2}$ , whose slope is $- \sqrt{2}$ , we have $- 2 \sin θ = - \sqrt{2} ⟹ \sin θ = \frac{1}{\sqrt{2}}$ . Since $P$ is in the first quadrant, $θ = \frac{π}{4}$ .
Execute: For $θ = \frac{π}{4}$ , the point of contact is $P (6 \sec \frac{π}{4}, 3 \tan \frac{π}{4}) = (6 \sqrt{2}, 3)$ . The equation of the normal becomes $6 x (\frac{1}{\sqrt{2}}) + 3 y (1) = 45 ⟹ 3 \sqrt{2} x + 3 y = 45 ⟹ \sqrt{2} x + y = 15$ . The $y$ -axis intercept $K$ is $(0, 15)$ .
Execute: The distance squared between $P (6 \sqrt{2}, 3)$ and $K (0, 15)$ is $d^{2} = (6 \sqrt{2} - 0)^{2} + (3 - 15)^{2} = 72 + (- 12)^{2} = 72 + 144 = 216$ .
Verify: Check that $P (6 \sqrt{2}, 3)$ lies on $H$ : $\frac{(6 \sqrt{2})^{2}}{36} - \frac{3^{2}}{9} = \frac{72}{36} - \frac{9}{9} = 2 - 1 = 1$ , which is correct. Normal at $(x_{1}, y_{1})$ has slope $- \frac{a^{2} y_{1}}{b^{2} x_{1}} = - \frac{36 \times 3}{9 \times 6 \sqrt{2}} = - \frac{108}{54 \sqrt{2}} = - \sqrt{2}$ , matching the given slope.

Length of a Normal Segment to a Hyperbola Intersecting the y-Axis

What feels right?

The value of d² is 216.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

08Execute

✓Verify

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