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Question

Question: The circle drawn with variable chord \[x+ay-5=0\] (\[a\] being parameter) of the parabola \[{{y}^{2}...

The circle drawn with variable chord x+ay5=0x+ay-5=0 (aa being parameter) of the parabola y2=20x{{y}^{2}}=20x as diameter will always touch the line
(a) x+5=0x+5=0
(b) y+5=0y+5=0
(c) x+y+5=0x+y+5=0
(d) xy+5=0x-y+5=0

Explanation

Solution

Hint: To find the circle with variable chord of the parabola as its diameter, we will

first find the exact equation of the chord and then use the fact that a circle drawn with a

chord to the parabola as its diameter has a tangent which is directrix to the parabola.We

have a variable chord x+ay5=0x+ay-5=0 of the parabolay2=20x{{y}^{2}}=20x.

We want to find the equation of tangent to the circle with the variable chord of the parabola

as its diameter.

We will consider the parabolay2=4ax{{y}^{2}}=4ax.

By comparing it with our parabolay2=20x=4×5x{{y}^{2}}=20x=4\times 5x, we observe thata=5a=5.

Substituting the value ofaain the equation of the chordx+ay5=0x+ay-5=0of the parabola, we

havex+5y5=0x+5y-5=0as the chord of the parabolay2=20x{{y}^{2}}=20x.

We know that a circle drawn with a chord to the parabola as its diameter has a tangent

which is directrix to the parabola.

We know that the equation of directrix to the parabolay2=4ax{{y}^{2}}=4axisx+a=0x+a=0.

Substitutinga=5a=5, we getx+5=0x+5=0as the equation of directrix to the parabolay2=20x{{y}^{2}}=20x.

Thus, the tangent to the circle drawn with chord of the parabolay2=20x{{y}^{2}}=20x as its diameter is of the formx+5=0x+5=0

Hence, the correct answer is x+5=0x+5=0.

Note: We can also solve the question by finding the exact equation of the circle and then finding the equation of tangent to the circle. To find the equation of the tangent to the

circle, we will find the equation of circle and points of intersection of the chord to the

parabola and the circle. However, it’s not necessary to do as it’s a longer way to solve the

question.