Question

The chord of the parabola $y=-a^2{x}^2+5ax-4$ touches the curve $y=\frac{1}{1-x}$ at the point $x=2$ and is bisected by that point. Find ${}^{\prime }{a}^{\prime }$.

Answer 1

Solution:-

$y=-a^2{x}^2+5ax-4$

$y=\frac{1}{1-x}$

Substituting $x=2$, we have

$y=\frac{1}{1-2}=-1$

Hence point of tangency is $(2,-1)$

$y=\frac{1}{1-x}$

Differentiating above equation w.r.t. $x$, we have

$\frac{dy}{dx}=-\frac{1}{{(1-x)}^2}\times (-1)=\frac{1}{{(1-x)}^2}$

At point of tangency, i.e., $(2,-1)$

$\frac{dy}{dx}=\frac{1}{{(1-2)}^2}=1$

$\therefore$ Equation of tangent is $[(y-y_1)=m(x-x_1)]$-

$y-(-1)=1(x-2)$

$\Rightarrow y-x+3=0.....\left(1\right)$

Now,

$y=-a^2{x}^2+5ax-4$

$\Rightarrow a^2{x}^2-5ax-y+4=0$

We know that equation of chord having mid point $(x_1,y_1)$ is-

$T=S_1$

$a^{2}x{x}_1-5a\left(\frac{x+x_1}{2}\right)+\left(\frac{y+y_1}{2}\right)+4=a^2{x_1}^2-5a{x}_1+y_1+4$

$a^{2}x{x}_1-5a\left(\frac{x+x_1}{2}\right)+\left(\frac{y+y_1}{2}\right)=a^2{x_1}^2-5a{x}_1+y_1$

Substituting $(x_1,y_1)=(2,-1)$ as the chord is bisected by the same point, we have

$2a^{2}x-5a\left(\frac{x+2}{2}\right)+\left(\frac{y+(-1)}{2}\right)=a^2{2}^2-10a+(-1)$

$2a^{2}x-\frac{5ax}{2}-5a+\frac{y}{2}-\frac{1}{2}=4a^2-10a-1$

$x(2a^2-\frac{5a}{2})+\frac{y}{2}=4a^2-5a-\frac{1}{2}$

$\Rightarrow x(4a^2-5a)+y=8a^2-10a-1.....\left(2\right)$

As equation $\left(1\right)\&\left(2\right)$ represents same line.

Therefore equating the corresponding coefficients, we have

coefficient of $x$ in equation $\left(2\right)=$ coefficient of $x$ in equation $\left(1\right)$

$4a^2-5a=-1$

$\Rightarrow 4a^2-5a+1=0$

$4a^2-4a-a+1=0$

$\Rightarrow 4a(a-1)-1(a-1)=0$

$\Rightarrow (4a-1)(a-1)=0$

$\Rightarrow a=\frac{1}{4}\text{or}1$

Hence the required answer is $a=\frac{1}{4}\text{or}1$.