Q11Written answer
If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.
Show solutionHide solution
Worked solution
NCERT Exemplar · Class 11 · Maths
Chapter 11: Conic Sections
Practice each question, then open the answer or worked solution.
Exercise · Page 202
Textbook page 202 · 28 questionsQ12Written answer
Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci.
Show solutionHide solution
Worked solution
Q13Written answer
If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse.
Show solutionHide solution
Worked solution
Q14Written answer
Find the equation of ellipse whose eccentricity is 2/3 , latus rectum is 5 and the centre is (0, 0).
Show solutionHide solution
Worked solution
Q15Written answer
Find the distance between the directrices of the ellipse x2/36 + y2/20 = 1
Show solutionHide solution
Worked solution
Q16Written answer
Find the coordinates of a point on the parabola y2 = 8x whose focal distance is 4.
Show solutionHide solution
Worked solution
We know that equation of an ellipse is y2 = 4ax,
Also we have length of latus rectum = 4a
Now by comparing the above two equations,
4a = 8
Therefore
a = 2
y2 = 8 × 2 = 16
y = ± 4 and x = 2
Hence the co – ordinates are (2, 4) and (2, -4).
Q17Written answer
Find the length of the line-segment joining the vertex of the parabola y2 = 4axand a point on the parabola where the line-segment makes an angle q to the x-axis.
Show solutionHide solution
Worked solution
Q18Written answer
If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.
Show solutionHide solution
Worked solution
Given
Q19Written answer
If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m.
Show solutionHide solution
Worked solution
Given equations are,
y = mx + 1 & y2 = 4x
By solving given equations we get
(mx + 1)2 = 4x
Expanding the above equation we get
m2x2 + 2mx + 1 = 4x
On rearranging we get
m2x2 + 2mx – 4x + 1 = 0
m+x2 + x (2m – 4) + 1 = 0
As the line touches the parabola, above equation must have equal roots,
Discriminant (D) = 0
(2m – 4)2 – 4 (m2) (1) = 0
4m2 – 16m + 16 – 4m2 = 0
-16 m + 16 = 0
– m + 1 = 0
m = 1
Hence, the required value of m is 1.
Q20Written answer
If the distance between the foci of a hyperbola is 16 and its eccentricity is √2, then obtain the equation of the hyperbola.
Show solutionHide solution
Worked solution
