Question

Find the equation of the tangent to the parabola $y=x^2-3x+2$ which is parallel to the line $5x-2y+10=0$. Find also the coordinates of the point of contact.

• A. $-\frac{2}{5}x$$-\frac{79}{50}$
• B. $-\frac{2}{5}x$$+\frac{79}{50}$
• C. $-\frac{4}{5}x$$-\frac{79}{50}$
• D. $-\frac{2}{5}x$$-\frac{9}{50}$

Answer 1

The correct option is A $-\frac{2}{5}x$$-\frac{79}{50}$

Slope of the tangent of the curve

$y=x^2-3x+2⟶\left(1\right)$

at a point $(h,k)$ as the curve is

$\frac{dy}{dx}=2h-3$

$\therefore$ equation of the tangent is

$y-k=(2h-3)(x-h)$

$\Rightarrow y-k=2hx-{2h}^2-3x+3h$

$\Rightarrow y=x(2h-3)-{2h}^2+3h+k⟶\left(2\right)$

$\Rightarrow 2h-3=\frac{5}{2}$

$\Rightarrow 2h=3+\frac{5}{2}=\frac{11}{2}$

$h=11/4$

Now $(h,k)$ lies on the curve $\left(1\right)$

$\Rightarrow K={\left(\frac{11}{4}\right)}^2-3\times \frac{11}{4}+2$

$=\frac{121}{16}-\frac{33}{4}+2$

$=\frac{121-132+32}{16}$

$=\frac{153-132}{16}$

$=\frac{21}{16}$

$\therefore pointofcontactis(11/4,21/16)$

And the equation of tangent is obtained by substituting

$h=11/4$ and $K=21/16$ in equation ##(2)$,

$y=x\left(\frac{5}{2}\right)-\frac{121}{8}+\frac{33}{4}+\frac{21}{16}$

$\Rightarrow y=\frac{5x}{2}-\frac{79}{16}$

(None of the options mentioned are correct)