The line x + y + 2 =0 is a tangent to a parabola $y^{2} = 4 a x$ at the point A. If it intersect the directrix at B and tangent at vertex at C, then

A B = \sqrt{32}

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A C = 4

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A C . B C = 8

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Area of

Δ A B C

= 5 sq. units.

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Solution

The correct options are
A $A B = \sqrt{32}$
C $A C . B C = 8$
$a m^{2} = l n \Rightarrow a = 2$
A = (2, -4), B = (-2, 0), C = (0, -2)
AB = $\sqrt{32}$
AC = $\sqrt{8}$ ,BC= $\sqrt{8}$
$∴$ AC .BC=8

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The line x + y + 2 =0 is a tangent to a parabola y2=4ax at the point A. If it intersect the directrix at B and tangent at vertex at C, then

The correct options are A AB=√32 C AC.BC=8am2=ln⇒a=2 A = (2, -4), B = (-2, 0), C = (0, -2) AB = √32 AC = √8 ,BC= √8 ∴ AC .BC=8

The line x + y + 2 =0 is a tangent to a parabola $y^{2} = 4 a x$ at the point A. If it intersect the directrix at B and tangent at vertex at C, then

The correct options are
A $A B = \sqrt{32}$
C $A C . B C = 8$
$a m^{2} = l n \Rightarrow a = 2$
A = (2, -4), B = (-2, 0), C = (0, -2)
AB = $\sqrt{32}$
AC = $\sqrt{8}$ ,BC= $\sqrt{8}$
$∴$ AC .BC=8