Question

The portion of the tangent to the curve $x=\sqrt{a^2-y^2}+\frac{a}{2}\log \frac{a-\sqrt{a^2-y^2}}{a+\sqrt{a^2-y^2}}$ intercepted between the curve and $x-$axis, is of length

• A. $\frac{a}{2}$
• B. $a$
• C. $2a$
• D. $\frac{a}{4}$

Answer 1

Answer: (B)

$a$

$x=\sqrt{a^2-y^2}+\frac{a}{2}\log \frac{a-\sqrt{a^2-y^2}}{a+\sqrt{a^2-y^2}}$ .....(1)

Differentiating w.r.t. $x$ , we get

$1=\frac{-y}{\sqrt{a^2-y^2}}\frac{dy}{dx}+\frac{ay}{2(a-\sqrt{a^2-y^2})\sqrt{a^2-y^2}}\frac{dy}{dx}+\frac{ay}{2(a+\sqrt{a^2-y^2})\sqrt{a^2-y^2}}\frac{dy}{dx}$

$\Rightarrow \frac{2\sqrt{a^2-y^2}}{y}=(-2+\frac{a}{(a-\sqrt{a^2-y^2})}+\frac{a}{(a+\sqrt{a^2-y^2})})\frac{dy}{dx}$

$\Rightarrow \frac{2\sqrt{a^2-y^2}}{y}=2\left(\frac{a^2-y^2}{y^2}\right)\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{y}{\sqrt{a^2-y^2}}$

Slope of tangent at $P(x_1,y_1)$ is $\frac{y_1}{\sqrt{a^2-y_1^2}}$

Equation of tangent at $P(x_1,y_1)$ is

$y-y_1=\frac{y_1}{\sqrt{a^2-y_1^2}}(x-x_1)$ ....(2)

Substituting $y=0$ in eqn (1), we get

$x=a$

So, the point A on axis is $(a,0)$

Since, the tangent (given by eqn (2)) passes through $(a,0)$

$\Rightarrow (x_1-a)=\sqrt{a^2-y_1^2}$ .....(3)

Now, portion of tangent between curve and x-axis is

$AP=\sqrt{(x_1-a{)}^2+y_1^2}$

$\Rightarrow AP=\sqrt{a^2}$ (by (3))

$\Rightarrow AP=a$

Hence, option B.