Exams Joint Entrance Examination Mathematics Trigonometric Ratios and Identities Question 103 Question 103 of 298 Medium NUMERICAL
Let E E E be an ellipse whose axes are parallel to the coordinate axes, having its centre at ( 3 , − 4 ) (3,-4) ( 3 , − 4 ) , one focus at ( 4 , − 4 ) (4,-4) ( 4 , − 4 ) and one vertex at ( 5 , − 4 ) (5,-4) ( 5 , − 4 ) . If m x − y = 4 mx-y=4 m x − y = 4 , m > 0 m>0 m > 0 is a tangent to the ellipse E E E , then the value of 5 m 2 5m^2 5 m 2 is equal to \_\_\_\_\_\_\_\_.
Answer: 3
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a = 2 a=2 a = 2 , a e = 1 ⇒ e = 1 2 ae=1\Rightarrow e=\dfrac12 a e = 1 ⇒ e = 2 1 , b 2 = a 2 − ( a e ) 2 = 4 − 1 = 3 b^2=a^2-(ae)^2=4-1=3 b 2 = a 2 − ( a e ) 2 = 4 − 1 = 3 .
\[E:\frac{(x-3)^2}4+\frac{(y+4)^2}3=1.\]
Substituting y = m x − 4 y=mx-4 y = m x − 4 : ( x − 3 ) 2 4 + m 2 x 2 3 = 1 ⇒ ( 3 + 4 m 2 ) x 2 − 18 x + 15 = 0 \dfrac{(x-3)^2}4+\dfrac{m^2x^2}3=1\Rightarrow(3+4m^2)x^2-18x+15=0 4 ( x − 3 ) 2 + 3 m 2 x 2 = 1 ⇒ ( 3 + 4 m 2 ) x 2 − 18 x + 15 = 0 .
Tangency: discriminant = 0 =0 = 0 : 324 − 4 ( 3 + 4 m 2 ) ( 15 ) = 0 ⇒ 144 − 240 m 2 = 0 ⇒ 5 m 2 = 3 324-4(3+4m^2)(15)=0\Rightarrow144-240m^2=0\Rightarrow5m^2=3 324 − 4 ( 3 + 4 m 2 ) ( 15 ) = 0 ⇒ 144 − 240 m 2 = 0 ⇒ 5 m 2 = 3 . Answer the question to unlock