मराठी

Find the area of the triangle formed by the lines joining the vertex of the parabola x 2 = 12 y to the ends of its latus rectum. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the area of the triangle formed by the lines joining the vertex of the parabola \[x^2 = 12y\]  to the ends of its latus rectum.

Advertisements

उत्तर

The given equation of the parabola is x2 = 12y.

On comparing the given equation with\[x^2 = 4ay\] 

a = 3

Required area = \[\frac{1}{2}\left( LL' \times OS \right) = \frac{1}{2} \times 12 \times 3 = 18 \text{ square units }\] 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 25: Parabola - Exercise 25.1 [पृष्ठ २५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
पाठ 25 Parabola
Exercise 25.1 | Q 6 | पृष्ठ २५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum.

y2 = 12x


Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum.

x2 = – 16y


Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum.

y2 = 10x


Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum.

x2 = –9y


Find the coordinates of the point of intersection of the axis and the directrix of the parabola whose focus is (3, 3) and directrix is 3x − 4y = 2. Find also the length of the latus-rectum. 


If b and c are lengths of the segments of any focal chord of the parabola y2 = 4ax, then write the length of its latus-rectum. 


(vii)  find the equation of the hyperbola satisfying the given condition:

foci (± 4, 0), the latus-rectum = 12


If the parabola y2 = 4ax passes through the point (3, 2), then find the length of its latus rectum. 


The vertex of the parabola (y + a)2 = 8a (x − a) is 


The length of the latus-rectum of the parabola y2 + 8x − 2y + 17 = 0 is 


The vertex of the parabola x2 + 8x + 12y + 4 = 0 is


The length of the latus-rectum of the parabola 4y2 + 2x − 20y + 17 = 0 is 


The length of the latus-rectum of the parabola x2 − 4x − 8y + 12 = 0 is 


The focus of the parabola y = 2x2 + x is 


Which of the following points lie on the parabola x2 = 4ay


The area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum is ______.


If the latus rectum of an ellipse with axis along x-axis and centre at origin is 10, distance between foci = length of minor axis, then the equation of the ellipse is ______.


If the eccentricity of an ellipse is `5/8` and the distance between its foci is 10, then find latus rectum of the ellipse.


If the parabola y2 = 4ax passes through the point (3, 2), then the length of its latus rectum is ______.


The length of the latus rectum of the ellipse 3x2 + y2 = 12 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×