Let $P$ be a point in the first quadrant that lies on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$.

Let $$a$$ and $$b$$ be positive numbers such that $$a>1$$ and $$b. Let $$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

(B) $$\sqrt{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 $$P$$.

• The point $(a^2,b^2)$ must satisfy the equation of the hyperbola. Therefore ... Commented Feb 24 at 9:29
• 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). Commented Feb 24 at 9:39
• @Dr.RichardKlitzing Got it, Thanks Commented Feb 24 at 11:09

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,

$$\Delta = b^4$$