Q&A - Ask Doubts and Get Answers

Sort by :
Clear All
Q

8.   An equilateral triangle is inscribed in the parabola y^2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Given, an equilateral triangle inscribed in parabola with the equation. The one coordinate of the triangle is A(0,0). Now, let the other two coordinates of the triangle are   and  Now, Since the triangle is equilateral, The coordinates of the points of the equilateral triangle are, So, the side of the triangle is   

6.  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.

Given the parabola, Comparing this equation with , we get Now, As we know the coordinates of ends of latus rectum are: So, the coordinates of latus rectum are, Now the area of the triangle with coordinates (0,0),(6,3) and (-6,3) Widht of the triangle = 2*6=12 Height of the triangle = 3 So The area =  Hence the required area is 18 unit square.

6.  A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Let  be the angle that rod makes with the ground, Now, at a point 3 cm from the end, At the point touching the ground Now, As we know the trigonometric identity, Hence the equation is,

4.   An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

The equation of the semi-ellipse will be of the form  Now, According to the question, the length of major axis = 2a = 8   The length of the semimajor axis =2 Hence the equation will be, Now, at point 1.5 cm from the end, the x coordinate is 4-1.5 = 2.5 So, the height at this point is  Hence the height of the required point is 1.56 m.

3.  The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m Find the length of a supporting wire attached to the roadway 18 m from the middle

Given, The width of the parabolic cable = 100m The length of the shorter supportive wire attached =  6m The length of the longer supportive wire attached = 30m Since the rope opens towards upwards, the equation will be of the form  Now if we consider origin at the centre of the rope, the equation of the curve will pass through points, (50,30-6)=(50,24) Hence the equation of the parabola...

2.   An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

Since the Axis of the parabola is vertical, Let the equation of the parabola be,  it can be seen that this curve will pass through the point (5/2, 10) if we assume origin at the bottom end of the parabolic arch. So, Hence, the equation of the parabola is  Now, when y = 2 the value of x will be  Hence the width of the arch at this height is 

1.   If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Le the parabolic reflector opens towards the right. So the equation of parabolic reflector will be, Now, Since this curve will pass through the point (5,10) if we assume origin at the optical centre, So Hence, The focus of the parabola is, . Alternative Method, As we know on any concave curve Hence, Focus  . Hence the focus is 5 cm right to the optical centre.

15. Find the equations of the hyperbola satisfying the given conditions.

       Foci (0,\pm\sqrt{10}), passing through (2,3)

Given, in a hyperbola,  Foci , passing through (2,3) Since foci of the hyperbola are in Y-axis, the equation of the hyperbola will be of the form ; By comparing standard parameter (foci) with the given one, we get Now As we know, in a hyperbola  Now As the hyperbola passes through the point (2,3) Solving Equation (1) and (2) Now, as we know that in a hyperbola  is always greater...

14.  Find the equations of the hyperbola satisfying the given conditions.

      vertices (± 7,0), e = \frac{4}{3}

Given, in a hyperbola vertices (± 7,0), And   Here, Vertices is  on the X-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (Vertices and eccentricity) with the given one, we get  and  From here, Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ;

13. Find the equations of the hyperbola satisfying the given conditions.

      Foci (± 4, 0), the latus rectum is of length 12

Given, in a hyperbola Foci (± 4, 0), the latus rectum is of length 12 Here,  focii are on the X-axis so, the standard equation of the Hyperbola will be ; By comparing standard parameter (length of latus rectum and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Since  can never be negative, Hence, The Equation of the hyperbola is ;

12.  Find the equations of the hyperbola satisfying the given conditions.

       Foci (\pm 3\sqrt5, 0), the latus rectum is of length 8.

Given, in a hyperbola Foci , the latus rectum is of length 8. Here,  focii are on the X-axis so, the standard equation of the Hyperbola will be ; By comparing standard parameter (length of latus rectum and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Since  can never be negative, Hence, The Equation of the hyperbola is ;

11.  Find the equations of the hyperbola satisfying the given conditions.

       Foci (0, ±13), the conjugate axis is of length 24.

Given, in a hyperbola Foci (0, ±13), the conjugate axis is of length 24. Here, focii are on the Y-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (length of conjugate axis and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ; .

10. Find the equations of the hyperbola satisfying the given conditions.

      Foci (± 5, 0), the transverse axis is of length 8.

Given, in a hyperbola Foci (± 5, 0), the transverse axis is of length 8. Here,  focii are on the X-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (transverse axis length and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ;

9.  Find the equations of the hyperbola satisfying the given conditions.

    Vertices (0, ± 3), foci (0, ± 5)

Given, in a hyperbola  Vertices (0, ± 3), foci (0, ± 5) Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (Vertices and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ; .

8.  Find the equations of the hyperbola satisfying the given conditions.

       Vertices (0, ± 5), foci (0, ± 8)

Given, in a hyperbola Vertices (0, ± 5), foci (0, ± 8) Here, Vertices and focii are on the Y-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (Vertices and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence, The Equation of the hyperbola is ; .

7. Find the equations of the hyperbola satisfying the given conditions. 

       Vertices (± 2, 0), foci (± 3, 0)

Given, in a hyperbola Vertices (± 2, 0), foci (± 3, 0) Here, Vertices and focii are on the X-axis so, the standard equation of the Hyperbola will be ; By comparing the standard parameter (Vertices and foci) with the given one, we get  and  Now, As we know the relation  in a hyperbola  Hence,The Equation of the hyperbola is ;

6.  Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

      49y^2 - 16x^2 = 784

Given a Hyperbola equation, Can also be written as   Comparing this equation with the standard equation of the hyperbola: We get,  and  Now, As we know the relation in a hyperbola, Therefore, Coordinates of the foci: The Coordinates of vertices: The Eccentricity: The Length of the latus rectum :

 5.  Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

     5y^2 - 9x^2 = 36

Given a Hyperbola equation, Can also be written as  Comparing this equation with the standard equation of the hyperbola: We get,    and  Now, As we know the relation in a hyperbola, Here as we can see from the equation that the axis of the hyperbola is Y-axis. So,  Coordinates of the foci: The Coordinates of vertices: The Eccentricity: The Length of the latus rectum :

 4.  Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

     16x^2 - 9y^2 = 576

Given a Hyperbola equation, Can also be written as  Comparing this equation with the standard equation of the hyperbola: We get,  and  Now, As we know the relation in a hyperbola, Therefore, Coordinates of the foci: The Coordinates of vertices: The Eccentricity: The Length of the latus rectum :

3.  Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

      9 y^2 - 4 x^2 =36

Given a Hyperbola equation, Can also be written as  Comparing this equation with the standard equation of the hyperbola: We get,  and  Now, As we know the relation in a hyperbola, Hence,  Coordinates of the foci: The Coordinates of vertices: The Eccentricity: The Length of the latus rectum :
Exams
Articles
Questions