NIMCET 2015 Question Paper with solutions

Question : 5 of 120

Marks: +1, -0

If 3x + 4y + K=0 is tangent to the hyperbola

9x^2

–

16y^2

= 144, then the value of K is;

Solution:

We have,

3x+4y+k=0

is tangent of the hyperbola

9x^2-16y^2=144

Equation of hyperbola is

9x^2-16y^2=144

⇒x^2/16-y^2/9=1

Equation of tangent is

3x+4y+k=0

We know that,

y=mx+c

touches the hyperbola.

\;\;{x^{2}}/{a^{2}}-\;\;{y^{2}}/{b^{2}}=1

c^{2}=a^{2}m^{2}-b^{2}

∴(\;\;{-k}/{4})^{2}=16(\;\;{-3}/{4})^{2}-9

⇒\;\;{k^{2}}/{16}=9-9

⇒k=0

Go to Question:

Prev Question

Next Question