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If two tangents drawn from a point (α, β) lying on the ellipse 25x2 + 4y2 = 1 to the parabola y2 = 4x are such that the slope of one tangent is four times the other, then the value of (10α + 5)2 + (16β2 + 50)2 equals ___________. (in integer)
Correct answer is '2929'. Can you explain this answer?
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Most Upvoted Answer
If two tangents drawn from a point (α, β) lying on the elli...
Called the external point) outside a circle are of equal length, then the external point lies on the circle's perpendicular bisector.
Proof:
Let O be the center of the circle, and let the tangents be AB and CD, with the external point P.
Since AB and CD are tangents to the circle, they are perpendicular to the radii OA and OC, respectively.
Therefore, ∠OAB = ∠OCD = 90°.
Also, since AB and CD are of equal length, AP = PD.
Therefore, triangle APD is an isosceles triangle, and so the perpendicular bisector of AD (which passes through P) also bisects APD.
Let this perpendicular bisector be represented by the line EF.
Since EF is a perpendicular bisector of AD, it passes through the midpoint of AD, which we can call M.
Therefore, EM = MF and ∠EMF = 90°.
Also, since OA and OC are radii of the circle, they are of equal length, and so OM = MF.
Therefore, EM = OM.
Therefore, triangle EOM is an isosceles triangle, and so ∠OEM = ∠EMO.
But ∠OAB = ∠OEM and ∠OCD = ∠EMO.
Therefore, ∠OAB = ∠OCD.
Therefore, P lies on the circle's perpendicular bisector.
Proof:
Let O be the center of the circle, and let the tangents be AB and CD, with the external point P.
Since AB and CD are tangents to the circle, they are perpendicular to the radii OA and OC, respectively.
Therefore, ∠OAB = ∠OCD = 90°.
Also, since AB and CD are of equal length, AP = PD.
Therefore, triangle APD is an isosceles triangle, and so the perpendicular bisector of AD (which passes through P) also bisects APD.
Let this perpendicular bisector be represented by the line EF.
Since EF is a perpendicular bisector of AD, it passes through the midpoint of AD, which we can call M.
Therefore, EM = MF and ∠EMF = 90°.
Also, since OA and OC are radii of the circle, they are of equal length, and so OM = MF.
Therefore, EM = OM.
Therefore, triangle EOM is an isosceles triangle, and so ∠OEM = ∠EMO.
But ∠OAB = ∠OEM and ∠OCD = ∠EMO.
Therefore, ∠OAB = ∠OCD.
Therefore, P lies on the circle's perpendicular bisector.
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Community Answer
If two tangents drawn from a point (α, β) lying on the elli...
∵
Tangent to the parabola y = mx + 1/m passes through (α, β). So, α m2 - βm + 1 = 0 has roots m1 and 4m1,
m1 + 4m1 = β/α and m1. 4m1 = 1/α
Gives: 4β2 = 25α ... (2)
From (1) and (2),
25(α2 + α) = 1 ... (3)
Now, (10α + 5)2 + (16β2 + 50)2
= 25(2α + 1)2 + 2500 (2α + 1)2
= 2525 (4α2 + 4α + 1) from equation (3)
= 2525 x
= 2929
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If two tangents drawn from a point (α, β) lying on the ellipse 25x2 + 4y2 = 1 to the parabola y2 = 4x are such that the slope of one tangent is four times the other, then the value of (10α + 5)2 + (16β2 + 50)2 equals ___________. (in integer)Correct answer is '2929'. Can you explain this answer?
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