Answer
451.2k+ views
Hint: We have to assume parabola which bisect the angle between the focal chord through P and perpendicular from P and perpendicular from P on the directrix.
Complete step-by-step answer:
Without loss of generality, Let’s assume the parabola is
${x^2} = 4ay$
The focus is (0,a) and the slope at any point $\left( {c,\dfrac{{{c^2}}}{{4a}}} \right)$ is $\dfrac{c}{{2a}}$ and the tangent equation is
$y = \dfrac{{{c^2}}}{{4a}} = \dfrac{c}{{2a}}\left( {x - c} \right)$
Let the distance d be
$d = \dfrac{{4a\left( a \right) - 2c\left( 0 \right) - {c^2} + 2{c^2}}}{{\sqrt {16{a^2} + 4{c^2}} }}$
Now let’s find its maximum
\[d = \dfrac{{4{a^2} + {c^2}}}{{\sqrt {16{a^2} + 4{c^2}} }}\]
$d = \dfrac{1}{2}\sqrt {4{a^2} + {c^2}} $
This distance has its maximum varying value of c at c=0
So d=a
Now we can say that perpendicular drawn from focus on any tangent to a parabola at any point lies on the tangent at vertex.
NOTE:
Whenever you come to this type of problem assume such a point on parabola which is mentioned above. By using this we can easily get the result that the foot of perpendicular (H) from the focus (S) on any tangent to a parabola at any point P lies on the tangent at vertex.
Complete step-by-step answer:
Without loss of generality, Let’s assume the parabola is
${x^2} = 4ay$
The focus is (0,a) and the slope at any point $\left( {c,\dfrac{{{c^2}}}{{4a}}} \right)$ is $\dfrac{c}{{2a}}$ and the tangent equation is
$y = \dfrac{{{c^2}}}{{4a}} = \dfrac{c}{{2a}}\left( {x - c} \right)$
Let the distance d be
$d = \dfrac{{4a\left( a \right) - 2c\left( 0 \right) - {c^2} + 2{c^2}}}{{\sqrt {16{a^2} + 4{c^2}} }}$
Now let’s find its maximum
\[d = \dfrac{{4{a^2} + {c^2}}}{{\sqrt {16{a^2} + 4{c^2}} }}\]
$d = \dfrac{1}{2}\sqrt {4{a^2} + {c^2}} $
This distance has its maximum varying value of c at c=0
So d=a
Now we can say that perpendicular drawn from focus on any tangent to a parabola at any point lies on the tangent at vertex.
NOTE:
Whenever you come to this type of problem assume such a point on parabola which is mentioned above. By using this we can easily get the result that the foot of perpendicular (H) from the focus (S) on any tangent to a parabola at any point P lies on the tangent at vertex.
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