Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.5
Question 1.
Find the equation of ellipse whose :
(i) Focus (- 1, 1), Directrix x – y + 4 = 0 and eccentricity is e – 1 / \(\sqrt { 5 }\)
(ii) Focus (- 2,3), Directrix 3x + 4y = 1 and eccentricity is e = 1/3
Solution:
(i) Let (h, k) be any point on ellipse, then according to definition,
Distance of P from focus = e(Distance of P from Directrix)
⇒ PS = e(PM)
⇒ (PS)2 = e2(PM)2
Thus locus of point P(h, k), 9x2 + 9y2 + xy – 16x – 16y + 4 = 0 which is required equation of ellipse.
(ii) Let P(h, k) be any point on ellipse, then according to definition
Distance of P from focus = e(Distance of P from directrix)
⇒ PS = e(PM)
⇒ (PS)2 = e2(PM)2
⇒ 225h2 + 225k2 + 900h – 1350k + 2925
– 9h2 – 16k2 – 1 – 24hk + 81 + 6h
⇒ 216h2 + 209k2 – 24hk + 906h – 1342k + 2924 = 0
At point (x, y).
216x2 + 209y2 – 24xy + 906x – 1342y + 2924 = 0
This is required equation.
Question 2.
Find the eccentricity, latus rectum and focus of the following ellipse :
(i) 4x2 + 9y2 = 1,
(ii) 25x2 + 4y2 = 100,
(iii) 3x2 + 4y2 – 12x – 8y + 4 = 0
Solution:
(i) Equation of ellipse,
4x2 + 9y2 = 1
Coordinate of focus coordinates of focus of ellipse will be (± ae, 0).
Question 3.
Find the equation of ellipse whose axis are coordinate axis and passes through points (6, 2) and (4, 3).
Solution:
Standard equation of ellipse
It passes through point (6, 2).
It also passes through point (4, 3).
Multiply eqn. (i) by 9 and eqn. (ii) by 4 then subtracting
Put the value of a2 in equation (i),
Question 4.
Find the eccentricity of ellipse whose latus rectum is half of its minor axis.
Solution:
Let equation of ellipse
\(\frac { { x }^{ 2 } }{ a^{ 2 } } \) + \(\frac { { y }^{ 2 } }{ b^{ 2 } } \) = 1
Question 5.
Find the locus of a point which moves such that sum of its distances from point (1, 0) and (- 1, 0) remains 3. Which curve is this locus ?
Solution:
Let P(h, k) is any point such that sum of whose distance from A(1, 0) and B(- 1, 0) remains 3.
According to questions,