A straight line parallel to the line 2x - y + 5 = 0 is also a tangent to the curve y² = 4x+5. Then the point of contact is

Correct Answer is : (−1,1)(-1,1)(−1,1)

Correct Answer is : (−1,1)(-1,1)(−1,1)

Test Index

KEAM 2014 Math Paper

Question : 32 of 120

Marks: +1, -0

2x - y + 5 = 0

y^2 = 4x+5

. Then the point of contact is

$(2,1)$
$(-1,1)$
$(1,3)$
$(3,4)$
$(-1,2)$

Solution:

Given equation of curve is

y^{2}=4 x+5

On differentiating both sides w.r.t.

x

, we get

2 y \;\frac{d y}{d x}=4

At point

(x_{1}, y_{1})

\;\frac{d y}{d x}=\;\frac{2}{y_{1}}=m_{1}

(say)

Also, given straight line is

2 x-y+5=0

So, the slope of this line is

m_{2}=2

since, the line is parallel to the tangent of the curve.

\therefore \;\; m_{1}=m_{2} \Rightarrow \frac{2}{Y_{1}}=2 \Rightarrow y_{1}=1

As the point

(x_{1}, y_{1})

lies on the given curve.

\therefore \;\; y_{1}^{2}=4 x_{1}+5

Put

y_{1}=1,

we get

1^{2}=4 x_{1}+5

\Rightarrow \;\; 4 x_{1}=-4

\Rightarrow \;\; x_{1}=-1

Hence, the point of contact is (-1,1)

Go to Question:

Prev Question

Next Question