JEE Advanced 2020 · Paper 2 · Mathematics Q8

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MathematicsMultiple correct+4 / -2Question 44 of 54

Let $a$ and $b$ be positive real numbers such that $a > 1$ and $b < a$ . Let $P$ be a point in the first quadrant that lies on the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ . Suppose the tangent to the hyperbola at $P$ passes through the point $(1, 0)$ , and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $\Delta$ denote the area of the triangle formed by the tangent at $P$ , the normal at $P$ and the $x$ -axis. If $e$ denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE?

1 < e < \sqrt2

\sqrt2 < e < 2

\Delta = a^4

\Delta = b^4

Answer and worked solution

Official answer: A, D

Let $P=(x_1,y_1)$ .

Tangent through $(1,0)$ . The tangent $\dfrac{xx_1}{a^2}-\dfrac{yy_1}{b^2}=1$ passes through $(1,0)$ , so $x_1=a^2$ .

Normal with equal intercepts. The normal has slope $-1$ , so the tangent has slope 1:
$\frac{b^2x_1}{a^2y_1}=1\implies y_1=b^2.$

$P$ on the hyperbola. $\dfrac{a^4}{a^2}-\dfrac{b^4}{b^2}=1$ , so $a^2-b^2=1$ .

(A)/(B) Eccentricity.
$e^2=1+\frac{b^2}{a^2}=2-\frac1{a^2}\in(1,2),$
so $1<e<\sqrt2$ . (A) is true and (B) is false.

(C)/(D) Area. The tangent $y=x-1$ meets the $x$ -axis at 1. The normal $y-b^2=-(x-a^2)$ meets it at $a^2+b^2$ . The base is $a^2+b^2-1=2b^2$ and the height is $y_1=b^2$ :
$\Delta=\tfrac12\cdot2b^2\cdot b^2=b^4.$
(D) is true and (C) is false.

(A), (D)

Topics:Hyperbola Tangents and normals to conics

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JEE Advanced 2020 · Paper 2, Mathematics question 8