JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The slope of the tangent to the curves

x = {t^2} + 3t - 8,\;y = 2{t^2} - 2t - 5

at the point

(2,\; - 1)\;

View answer and explanation

Correct option: B

\frac{6}{7}

Option B (

\frac{6}{7}

) is correct because it is the slope of the tangent calculated at the given point by finding dy/dx using parametric differentiation.

More Questions from TANGENTS & NORMALS

Practice All Chapter MCQs →

a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z