Question

Let a(3,2) and b (4,7) are the foci of an ellipse and the line x+y-2=0 is a tangent to the ellipse, then the point of contact of this tangent with the ellipse is

Answer 1

We can also prove the above result by using the fact that the line = mx + √(7m2+3) will be tangent to x2 + y2 = 4 if discriminant of x2 + (m + √(7m2 + 3))2 = 4 is zero. Let P and Q be the points of contact of the common tangent with ellipse and circle respectively and O be the common centre of the two, then. PQ = √(OP2 ...

Answer 2

Answer:

Let a(3,2) and b (4,7) are the foci of an ellipse and the line x+y-2=0 is a tangent to the ellipse, then the point of contact of this tangent with the ellipse is (1,1).

Step-by-step explanation:

Let the points are x1 , y1 .

We know the distance of the point = $\sqrt{(x-3)^{2} + (y-2)^{2} } + \sqrt{(x-4)^{2} + (y-7)^{2} } = 4\sqrt{5}\\\\ or , \sqrt{(x-3)^{2} + (y-2)^{2} } = 4\sqrt{5} - \sqrt{(x-4)^{2} + (y-7)^{2} } \\\\or , (x-3)^{2} + (y-2)^{2} = 80 + (x-4)^{2} + (y-7)^{2} - 8\sqrt{5} \sqrt{(x-4)^{2} + (y-7)^{2} } \\\\or , x^{2} -6x+9 + y^{2} -4y+4 = 80 + x^{2} -8x+16+y^{2} -14y+49-8\sqrt{5} \sqrt{(x-4)^{2} + (y-7)^{2} } \\\\or , 2x+10y-132 = -8\sqrt{5} \sqrt{(x-4)^{2} + (y-7)^{2} } \\\\or ,x+5y-66 = -4\sqrt{5} \sqrt{(x-4)^{2} + (y-7)^{2} }$

Now the equation of the line is

$x+y-2=0\\or , y=2-x$.......... (i)

This equation satisfy the previous equation

$x + 5(2-x)-66 = -4\sqrt{5}\sqrt{(x-4)^{2} +(2-x-7)^{2} } \\\\or , -4x-56 = -4\sqrt{5}\sqrt{x^{2} -8x+16+x^{2} +10x+25}\\ \\or , 2x+28= 2\sqrt{5}\sqrt{2x^{2} +2x+41}\\ \\ or ,x+14 = \sqrt{5}\sqrt{2x^{2} +2x+41}\\\\or ,x^{2} +28x+196 = 10x^{2} +10x+205\\ \\or , 9x^{2} -18x+9 = 0\\\\or , x^{2} -2x+1 =0\\\\or , (x-1)^{2} =0.\\\\\\x=1\\y=2-x=1.$

Let a(3,2) and b (4,7) are the foci of an ellipse and the line x+y-2=0 is a tangent to the ellipse, then the point of contact of this tangent with the ellipse is (1,1).

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