JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The slope of tangent to the curve

x=t^2+3 t-8

y=2 t^2-2 t-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 value of

\frac{dy}{dx}

at t=2, which is found by solving for t and then substituting into the derivative.

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