The value of $a$ for which the curves $y^{2} = 4 x$ and $x^{2} + y^{2} - 2 a x = 0$ intersect exactly at one point is

1

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Solution

The correct option is C $2$
Substituting $y^{2} = 4 x$ in $x^{2} + y^{2} - 2 a x = 0$ gives
$x^{2} + 4 x - 2 a x = 0$
$\Rightarrow x^{2} + (4 - 2 a) x = 0$
For this quadratic equation to have only one solution, discriminant has to be equal to zero.
$\Rightarrow (4 - 2 a)^{2} = 0$
$\Rightarrow 4 - 2 a = 0$
$\Rightarrow a = 2$

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The value of a for which the curves y2=4x and x2+y2−2ax=0 intersect exactly at one point is

The correct option is C 2Substituting y2=4x in x2+y2−2ax=0 gives x2+4x−2ax=0 ⇒x2+(4−2a)x=0 For this quadratic equation to have only one solution, discriminant has to be equal to zero. ⇒(4−2a)2=0 ⇒4−2a=0 ⇒a=2

The value of $a$ for which the curves $y^{2} = 4 x$ and $x^{2} + y^{2} - 2 a x = 0$ intersect exactly at one point is