Question

Find the point on the ellipse (6x²+ 11
${y}^{2}$
- 256.
where the common tangent to it and the
circle x² + y²-2x=15 touch​

Answer 1

Answer:

4x+3y+19=0

4x+3y−31=0

Step-by-step explanation:

We know that from the standard equation of the circle,one can find centre and radius of the circle

\begin{gathered} {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0 \\ \\ \end{gathered}

x

2

+y

2

+2gx+2fy+c=0

center (-g,-f) and radius

\begin{gathered}r = \sqrt{ {g}^{2} + {f}^{2} - c} \\ \\ \end{gathered}

r=

g

2

+f

2

−c

So,in the given equation

\begin{gathered} {x}^{2} + {y}^{2} - 6x + 4y - 12 = 0 \\ \\ center(3 ,- 2) \\ \\ r = \sqrt{9 + 4 + 12} \\ \\ = \sqrt{25} \\ \\ r = 5 \\ \\ \end{gathered}

x

2

+y

2

−6x+4y−12=0

center(3,−2)

r=

9+4+12

=

25

r=5

We also know that slope of parallel lines are equal and tangent touches the circle and make 90° angle with radius of the circle.

Hence equation of tangent would be

\begin{gathered}4x + 3y + k = 0 \\ \end{gathered}

4x+3y+k=0

and perpendicular distance from center will equal to the radius of circle,so

apply the distance formula between a line and a point

\begin{gathered} \bigg| \frac{ax + by + c}{ \sqrt{ { {x}^{2} } + {y}^{2} } } \bigg| \\ \\ \bigg| \frac{4(3) +( - 2)(3) + k}{ \sqrt{ { {4}^{2} } + {3}^{2} } } \bigg| = 5 \\ \\ \bigg| \frac{12 - 6 + k }{ \sqrt{25}} \bigg| = 5\\ \\ 6 + k =± 25 \\ \\ k = 25 - 6 \\ \\ k = 19 \\ \\or \\\\k=-25-6=-31 \end{gathered}

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x

2

+y

2

ax+by+c

∣

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4

2

+3

2

4(3)+(−2)(3)+k

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=5

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25

12−6+k

∣

∣

∣

∣

∣

=5

6+k=±25

k=25−6

k=19

or

k=−25−6=−31

Hence equation of tangent is

\begin{gathered}4x + 3y + 19 = 0 \\\\4x + 3y -31 = 0 \end{gathered}

4x+3y+19=0

4x+3y−31=0

Hope it helps you.