Given : A circle, 2x^{2}+2y^{2}=5  and a parabola, y^{2}=4\sqrt{5}x.

Statement - I : An equation of a common tangent to these curves is y=x+ \sqrt{5}.

Statement - II : If the line ,y=mx+\frac{\sqrt{5}}{m}(m\neq 0) is their common tangent, then m satisfies m^{4}-3m^{2}+2=0.

 

  • Option 1)

    Statement - I is false ; Statement - II is true.

     

  • Option 2)

    Statement - I is ture ; Statement - II is true ; Statement - II is a correct explanation for statement - I.

     

     

  • Option 3)

    Statement - I is ture ; Statement - II is true ; Statement - II is not a correct explanation for statement - I.

     

     

  • Option 4)

    Statement - I is ture ; Statement - II is false.

     

 

Answers (2)

As we learnt in

Condition of tangency -

c^{2}=a^{2}\; (1+m^{2})

 

- wherein

If  y=mx+c  is a tangent to the circle x^{2}+y^{2}=a^{2}

 

and

 

Equation of tangent -

y= mx+\frac{a}{m}

- wherein

Tengent to y^{2}=4ax is slope form.

 

x^{2}+y^{2}=\frac{5}{2}\:and\:y^{2}=4\sqrt{5x}

Tangent to circle is y=mx+\frac{5}{2}\sqrt{1+m^{2}}

Tangent to parabola is y=mx+\frac{\sqrt{5}}{m}

So, \frac{5}{2}\sqrt{1+m^{2}}=\frac{\sqrt{5}}{m}

On solving m=1

Thus tangent is y=x+\sqrt{5} 

 

 


Option 1)

Statement - I is false ; Statement - II is true.

 

This option is incorrect.

Option 2)

Statement - I is ture ; Statement - II is true ; Statement - II is a correct explanation for statement - I.

 

 

This option is incorrect.

Option 3)

Statement - I is ture ; Statement - II is true ; Statement - II is not a correct explanation for statement - I.

 

 

This option is correct.

Option 4)

Statement - I is ture ; Statement - II is false.

 

This option is incorrect.

N neha

Preparation Products

JEE Main Rank Booster 2021

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

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

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

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

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

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

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

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

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

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions