
The condition that the two tangents to the parabola \[{{y}^{2}}=4ax\]become normal to the circle \[{{x}^{2}}+{{y}^{2}}-2ax-2by+c=0\]is given by
(a). \[{{a}^{2}}>4{{b}^{2}}\]
(b). \[{{b}^{2}}>2{{a}^{2}}\]
(c) .\[{{a}^{2}}>2{{b}^{2}}\]
(d). \[{{b}^{2}}>4{{a}^{2}}\]
Answer
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Hint: The equation of tangents to the parabola becomes normal to the circle when the equation of tangents pass through the centre of the circle.
Complete step-by-step answer:
We have a parabola \[{{y}^{2}}=4ax\] and a circle \[{{x}^{2}}+{{y}^{2}}-2ax-2by+c=0\] .We want the equation of tangents of the parabola to be the equation of normal to the circle.
Consider any point on the parabola \[{{y}^{2}}=4ax\] of the form \[(a{{t}^{2}},2at)\].
We know that the equation of tangent of the parabola of the form \[{{y}^{2}}=4ax\] at a point\[(a{{t}^{2}},2at)\]is\[ty=x+a{{t}^{2}}\].
As the tangent of the parabola is normal to the circle, the equation of tangent must pass through the centre of the circle.
We know that the centre of circle of the form \[{{x}^{2}}+{{y}^{2}}-2ax-2by+c=0\]is\[(a,b)\].
Substituting the point \[(a,b)\] in the equation of tangent \[ty=x+a{{t}^{2}}\], we have \[tb=a+a{{t}^{2}}\].
We need to find real roots of the above quadratic equation to make the equation of tangent pass through the centre of the circle.
We know that the quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] has real roots when \[{{b}^{2}}-4ac>0\]
So, the equation \[tb=a+a{{t}^{2}}\]has real roots when \[{{b}^{2}}-4{{a}^{2}}>0\].
Thus, to make the equation of tangent pass through the centre of parabola( or to make the equation of the tangent same as the equation of normal to the circle), it is necessary that \[{{b}^{2}}-4{{a}^{2}}>0\].
Hence, the correct answer is \[{{b}^{2}}>4{{a}^{2}}\]
Note: We can also solve this question by writing the equation of normal to the circle passing through the centre of the circle with given coordinates and then comparing it with the tangent of the parabola to get the necessary condition. Also, we can write the equation of tangent to the parabola in one-point form instead of slope form.
Complete step-by-step answer:
We have a parabola \[{{y}^{2}}=4ax\] and a circle \[{{x}^{2}}+{{y}^{2}}-2ax-2by+c=0\] .We want the equation of tangents of the parabola to be the equation of normal to the circle.
Consider any point on the parabola \[{{y}^{2}}=4ax\] of the form \[(a{{t}^{2}},2at)\].
We know that the equation of tangent of the parabola of the form \[{{y}^{2}}=4ax\] at a point\[(a{{t}^{2}},2at)\]is\[ty=x+a{{t}^{2}}\].
As the tangent of the parabola is normal to the circle, the equation of tangent must pass through the centre of the circle.
We know that the centre of circle of the form \[{{x}^{2}}+{{y}^{2}}-2ax-2by+c=0\]is\[(a,b)\].
Substituting the point \[(a,b)\] in the equation of tangent \[ty=x+a{{t}^{2}}\], we have \[tb=a+a{{t}^{2}}\].
We need to find real roots of the above quadratic equation to make the equation of tangent pass through the centre of the circle.
We know that the quadratic equation of the form \[a{{x}^{2}}+bx+c=0\] has real roots when \[{{b}^{2}}-4ac>0\]
So, the equation \[tb=a+a{{t}^{2}}\]has real roots when \[{{b}^{2}}-4{{a}^{2}}>0\].
Thus, to make the equation of tangent pass through the centre of parabola( or to make the equation of the tangent same as the equation of normal to the circle), it is necessary that \[{{b}^{2}}-4{{a}^{2}}>0\].
Hence, the correct answer is \[{{b}^{2}}>4{{a}^{2}}\]
Note: We can also solve this question by writing the equation of normal to the circle passing through the centre of the circle with given coordinates and then comparing it with the tangent of the parabola to get the necessary condition. Also, we can write the equation of tangent to the parabola in one-point form instead of slope form.
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