Find the equations of the tangent and normal to the parabola y^2=4ax at the point  \left ( at^2,2at \right ). 

 

 

 

 
 
 
 
 

Answers (1)

Given curve is y^2=4ax

We need to find equation of tangent and normal at \left ( at^2,4at \right ).

We know that slope of tangent is  \frac{\mathrm{d} y}{\mathrm{d} x}.

y^2=4ax

Differentiating wrt x,

\frac{\mathrm{d} (y^2)}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}(4ax)

\frac{\mathrm{d} (y^2)}{\mathrm{d} y}\times \frac{\mathrm{d} y}{\mathrm{d} x}=4a\frac{\mathrm{d}(x) }{\mathrm{d} x}

2y\times \frac{\mathrm{d} y}{\mathrm{d} x}=4a

\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{4a}{2y}

Slope of tangent at  (at^2,2at)  is 

\left [ \frac{\mathrm{d} y}{\mathrm{d} x} \right ]_{(at^2,2at)}=\frac{4a}{2(2at)}=\frac{4a}{4at}=\frac{1}{t}

Also we know that 

Slope of tangent \times Slope of normal =-1

\frac{1}{t}\times sl\!ope\: \: o\!f\: \: normal=-1

\therefore sl\!ope\: \: o\!f\: \: normal=-t

 

 

 

Preparation Products

Knockout NEET July 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Test Series NEET July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 11999/-
Buy Now
Exams
Articles
Questions