Let and be positive real numbers such that and Let be a point in the first quadrant that lies on the hyperbola Suppose the tangent to the hyperbola at passes through the point (1, 0) and suppose the normal to the hyperbola at cuts off equal interc

Question

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 $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 .$ 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 $Δ$ 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? [2020]

Smart Achievers · Accepted Answer

(1, 4)

Let $P (a \sec θ, b \tan θ)$

Equation of tangent at $P$

$\frac{x}{a} \sec θ - \frac{y}{b} \tan θ = 1$

$∵$ $(1, 0) lies on the tangent$

so $a = \sec θ$

Equation of normal at $P$

$\frac{a x}{\sec θ} + \frac{b y}{\tan θ} = a^{2} + b^{2}$

since normal at $P$ makes equal intercept on co-ordinate axes,

$∴$ $Slope of normal is - 1$

so $- \frac{a}{b} \sin θ = - 1$

$\Rightarrow b = \tan θ,$ hence, $a^{2} - b^{2} = 1 ...(i)$

$∵$ $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$ $= \sqrt{1 + \frac{a^{2} - 1}{a^{2}}}$ $= \sqrt{2 - \frac{1}{a^{2}}} from (i)$

Since $a > 1,$ so $e \in (1, \sqrt{2})$

Hence, option (1) is true.

Area of $∆ P A B$ $= \frac{1}{2} A P \cdot P B$

$= \frac{1}{2} \sqrt{{(a^{2} - 1)}^{2} + {(b^{2})}^{2} \times \sqrt{2 b}}$ ⁴

$= \frac{1}{2} \sqrt{2 b^{4}} \sqrt{2 b^{4}}$ $= b^{4} from (i)$

Hence, option (4) is true.

Solution

1 $1 < e < \sqrt{2}$

2 $\sqrt{2} < e < 2$

3 $Δ = a^{4}$

4 $Δ = b^{4}$

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Solution

1 1<e<2

2 2<e<2

3 Δ=a4

4 Δ=b4

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1 $1 < e < \sqrt{2}$

2 $\sqrt{2} < e < 2$

3 $Δ = a^{4}$

4 $Δ = b^{4}$