The point on the curve y=5+x-x² at which the normal makes equal intercepts is

Correct Answer is : (0,5)(0,5)(0,5)

Correct Answer is : (0,5)(0,5)(0,5)

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KEAM 2013 Math Paper

Question : 102 of 120

Marks: +1, -0

y=5+x-x^2

at which the normal makes equal intercepts is

$(1,5)$
$(0,-1)$
$(-1,3)$
$(0,3)$
$(0,5)$

Solution:

Given curve is

y=5+x-x^{2}

.....(i)

On differentiating w.r.t.

x

, we get

\;\frac{d y}{d x}=1-2 x

Slope of normal

=\;\frac{-1}{\frac{d y}{d x}}=\;\frac{-1}{1-2 x}

=\;\frac{1}{2 x-1}

......(ii)

since, normal makes equal intercepts.

\therefore \theta=135^{\circ}

From Eq. (ii),

\;\frac{1}{2 x-1}=\tan 135^{\circ}=-1

\Rightarrow \;\; -2 x+1=1 \Rightarrow x=0

Then, from Eq. (i),

y=5

So, the required point is (0,5)

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