JEE Advanced 2015 · Paper 1 · Mathematics Q45

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MathematicsInteger+4Question 45 of 60

If the normals of the parabola $y^2=4x$ drawn at the end points of its latus rectum are tangents to the circle $(x-3)^2+(y+2)^2=r^2$ , then the value of $r^2$ is

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Answer and worked solution

Official answer: 2

Normals at the latus-rectum ends (1, ±2).

At (1, 2), the slope is −1: $x+y=3$ .
At (1, −2), the slope is 1: $x-y=3$ .

Tangency. The distance from $(3,-2)$ to $x+y-3=0$ is $\dfrac{|3-2-3|}{\sqrt2}=\sqrt2$ , and it is the same for the other line. So $r^2=2$ .

Answer: 2

Topics:Tangents and normals Tangents and normals to conics

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JEE Advanced 2015 · Paper 1, Mathematics question 45