The equation of the tangent line to the curve $y = x^{2} - 2 x + 7$ which is perpendicular to the line $5 y - 15 x = 13$ is $12 x + 36 y - 227 = 0$ .If true enter 1 else 0.

Open in App

Solution

Differentiating the equation of the curve with respect to $x$ , we get
$\frac{d y}{d x} = 2 x - 2$
Slope of the line $5 y - 15 x = 13$ is $3$ .
Since the tangent is perpendicular to the above line,
$\frac{d y}{d x} \times 3 = - 1$
$\Rightarrow \frac{d y}{d x} = \frac{- 1}{3}$
Using in the first equation, we get $\frac{- 1}{3} = 2 x - 2$
$\Rightarrow x = \frac{5}{6}$
Substituting the value of $x$ in the equation of the curve, we get
$y = {(\frac{5}{6})}^{2} - 2 (\frac{5}{6}) + 7$
$\Rightarrow y = \frac{217}{36}$
Hence, equation of the tangent is
$y - \frac{217}{36} = \frac{- 1}{3} (x - \frac{5}{6})$
$\Rightarrow 36 y - 217 = - 2 (6 x - 5)$
$\Rightarrow 12 x + 36 y - 227 = 0$

Suggest Corrections

Similar questions

Find the equation of the tangent line to the curve

y = x^{2} - 2 x + 7

which is perpendicular to the line

5 y - 15 x = 13.

Q. Find the equation of the tangent line to the curve y = x² − 2x + 7 which is (i) parallel to the line 2x − y + 9 = 0 (ii) perpendicular to the line 5y − 15x = 13.

Find the equation of the tangent line to the curve $y = x^{2} - 2 x + 7$ which is

(a) parallel to the line 2x-y+9=0.

(a) parallel to the line 5y-15x=13.

The equation of the straight line passing through the point of intersection of lines 3x – 4y – 7 = 0 and 12x – 5y – 13 = 0 and perpendicular to the line 2x – 3y + 5 = 0 is

Q. Equations of tangents perpendicular to the line

2 x + 3 y + 7 = 0

to the circle

x^{2} + y^{2} - 2 x + 4 y + 1 = 0

are :

Join BYJU'S Learning Program

The equation of the tangent line to the curve y=x2−2x+7 which is perpendicular to the line 5y−15x=13 is 12x+36y−227=0.If true enter 1 else 0.

The equation of the tangent line to the curve $y = x^{2} - 2 x + 7$ which is perpendicular to the line $5 y - 15 x = 13$ is $12 x + 36 y - 227 = 0$ .If true enter 1 else 0.