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Let $a$ and $b$ be positive 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 hyperbola at $P$ passes through the point $(1,0)$, and suppose the normal to hyperbola at $P$ cuts off equal intercepts on the coordinate axes. let $\triangle$ denotes the area of triangle formed by tangent at $P$, the normal at $P$ and $x$-axis. If $e$ denotes the eccentricity of the hyperbola, then which of the following statements is\are true?

(A) $1<e<\sqrt{2}$

(B) $\sqrt{2}<e<2$

(C) $\triangle = a^4$

(D) $\triangle = b^4$

Let parametric coordinate of P on Hyperobla as $(a \sec{\theta}, b\tan{\theta})$

Equation of tangent to hyperbola in parametric form be as $\dfrac{x}{a} \sec{\theta}-\dfrac{y}{b} \tan{\theta}=1$ and this tangent passes through $(1,0)$ , so I obtained $a=\sec{\theta}$

Also normal make equal intercept on coordinate axes of slope of normal will be either $1$ or $-1$ but since $P$ lies in first quadrant its value will be $-1$.

Now Equation of normal in parametric form for hyperobla is given by $ax \cos{\theta}+by \cot{\theta}=a^2+b^2$ and slope of normal is $\dfrac{-a}{b}\sin{\theta}$ which is equal to $-1$ so I obtained $b=a \sin{\theta}$

As I obtained $a=\sec{\theta}$ and $b=a \sin{\theta}\implies b=\tan{\theta}$

So point $P$ will be $(a^2, b^2)$ but this point does not satisfy equation of Hyperbola.

Am i missing anything?

This question was aksed in JEE Adavnaced 2020 Paper 2. And all the insitituation in India gave answer as $A$ and $D$

I am also getting answer as $A$ and $D$ but I have doubt about Point $P$.

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  • $\begingroup$ The point $(a^2,b^2)$ must satisfy the equation of the hyperbola. Therefore ... $\endgroup$
    – Lozenges
    Commented Feb 24 at 9:29
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    $\begingroup$ You probably was mislead by confusing $(a^2, b^2)$ with $(a, b)$. The latter indeed lies on the asymptote, not on the hyperbola. OTOH, $x_0=a^2$ inserted into the hyperbola yields $y_0=b \sqrt{a^2-1}$, thus your other solution value $y_0=b^2$ just imposes the thereby derived dependency $a^2=b^2+1$, i.e. $e^2=2-1/a^2$, implying indeed (A). $\endgroup$ Commented Feb 24 at 9:39
  • $\begingroup$ @Dr.RichardKlitzing Got it, Thanks $\endgroup$
    – mathophile
    Commented Feb 24 at 11:09

1 Answer 1

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Let $P = (x_1,y_1)$ lie on the hyperbola. The normal vector at $P$ is along

$ g = [ \dfrac{x_1}{a^2} , -\dfrac{y_1}{b^2} ] $

Therefore the equation of tangent line at $P$ is

$ \dfrac{x_1 (x - x_1)}{a^2} - \dfrac{ y_1 (y - y_1) }{b^2} = 0 $

Since $(1,0)$ lies on this tangent, then

$ \dfrac{ x_1 (1 - x_1) }{a^2} + \dfrac{ y_1^2}{b^2} = 0 $

But,

$ \dfrac{ x_1^2 }{a^2} - \dfrac{ y_1^2}{b^2} = 1 $

Therefore,

$ \dfrac{x_1}{a^2} = 1 $

Hence, $x_1 = a^2$

Substituting this into the hyperbola equation,

$ a^2 = 1 + \dfrac{y_1^2}{b^2} $

$ a^2 b^2 = b^2 + y_1^2 $

So that,

$ y_1^2 = b^2 (a^2 - 1 ) $

i.e. $ y_1 = b \sqrt{a^2 - 1} $

The equation of the normal line to the hyperbola at $P$ is

$ \dfrac{y_1 (x - x_1)}{b^2} + \dfrac{ x_1 (y - y_1) }{a^2} = 0 $

It's intersection with the $x-axis$ is at

$ \dfrac{y_1 (x - x_1)}{b^2} = \dfrac{ x_1 y_1 }{a^2} $

So,

$ a^2 y_1 (x - x_1) = b^2 x_1 y_1 $

$ a^2 x y_1 = x_1 y_1 (a^2 + b^2) $

$ x = \left( 1 + \left(\dfrac{b}{a}\right)^2 \right) x_1 = a^2 + b^2 $

And this is the $x$-intercept. Similarly, for the $y$ intercept

$ \dfrac{y_1 (- x_1)}{b^2} + \dfrac{ x_1 (y - y_1) }{a^2} = 0 $

So that

$ a^2 (- x_1 y_1) + b^2 x_1 (y - y_1) = 0 $

$ b^2 x_1 y = x_1 y_1 (a^2 + b^2) $

So that the the $y$ intercept is

$ y = y_1 ( (a/b)^2 + 1) = b \sqrt{a^2 - 1} ( (a/b)^2 + 1 ) $

Since the two intercepts are equal then

$ a^2 + b^2 = b \sqrt{a^2 - 1} ( (a/b)^2 + 1 ) $

$ (a^2 + b^2)^2 = b^2 (a^2 - 1) (a^2 + b^2)^2 / b^4 $

$ b^2 = a^2 - 1 $

Hence,

$ a^2 - b^2 = 1$

So now the indicated triangle has vertices:

$(1,0), (a^2 + b^2, 0) , (a^2 , b^2 )$

The area is given by

$ \Delta = \frac{1}{2} (a^2 + b^2 - 1) b^2 $

but $a^2 - 1 = b^2$ Therefore,

$ \Delta = b^4 $

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