Let the quadratic curve passing through the point (- 1, 0) and...

Given,

The quadratic curve passing through the point $\left(- 1, 0\right)$ and touching the line $y = x$ at \left(1, 1\right)$be$y = f\left(x\right)$,</p><p>Now let the curve be$y = a{x}^{2} + bx + c$</p><p>Passes through \left(- 1, 0\right)$

$⇒ a - b + c = 0 . . . . \left(i\right)$

Passes through \left(1, 1\right)$</p><p>$⇒ a + b + c = 1 . . . . . \ldots \left(ii\right)$</p><p>Now given tangent to$y = x$at \left(1, 1\right)$,

So, $\dfrac{dy}{dx} = 1$

$⇒ 2a + b = 1 . . . . \ldots \left(iii\right)$

Solving all above three equations we get,

$a = \dfrac{1}{4}, b = \dfrac{1}{2}$ and $c = \dfrac{1}{4}$

Hence, the equation of curve is $y = \dfrac{{x}^{2}}{4} + \dfrac{x}{2} + \dfrac{1}{4}$

Now \left(\alpha, \alpha + 1\right)$lies on curve</p><p>$\alpha +1= \dfrac{{\alpha}^{2}}{4} + \dfrac{\alpha}{2} + \dfrac{1}{4}$</p><p>$⇒ ( \alpha - 3 ) ( \alpha + 1 ) = 0$</p><p>$⇒ \alpha = 3$<br />Point in first quadrant is \left(3, 4\right)$

Now finding slope of tangent at \left(3, 4\right)$we get,</p><p>${\left(\dfrac{dy}{dx}\right)}_{\left(3, 4\right)} = {\left(\dfrac{x}{2} + \dfrac{1}{2}\right)}_{\left(3, 4\right)} = 2$</p><p>$⇒$Slope of normal$= \dfrac{- 1}{2}$</p><p>Hence, equation of normal will be,$y - 4 = - \dfrac{1}{2} ( x - 3 )$</p><p>So,$x$- intercept$= 11