Suppose, 2 x cos ⁡ α − 3 y sin ⁡ α = 6 is equation of a variable straight line. From two point A ( 5 , 0 and B ( − 5 , 0 ) foot of the altitude on that straight line is P and Q respectively. Show that the product of the length of two line segment AP and BQ is free of α .

Question

Suppose,   2  x  cos  ⁡  α  −  3  y  sin  ⁡  α  =  6 is equation of a variable straight line. From two point   A  (      5    ,  0 and   B  (  −      5    ,  0  ) foot of the altitude on that straight line is P and Q respectively. Show that the product of the length of two line segment AP and BQ is free of   α .

marktje28 · Accepted Answer

Step 1      2    x    cos    ⁡    α    −    3    y    sin    ⁡    α    =    6        ⟹        y    =                  2        cot        ⁡        α            3        x    −    2    csc    ⁡    α    =    m    x    +    c    where   m  =  tan  ⁡  θ  =                    2        cot        ⁡        α            3      Consider   0  ≤  α  ≤  2  π that gives all possible lines for the given equation. If   α  =      π    2   or             3      π        2  , the line is   y  =  −  2 or   y  =  2 respectively and the perp distance from both points on x-axis would be 2. So the product is 4. For other values of   αThe line intersects x-axis at           x    =          3              cos        ⁡        α            So distance of the given points to the intersection point is,      |                  5            ±              3                  cos          ⁡          α                      |  The product of perpendicular lengths to the line is then given by,            |      5      −              9                              cos            2                    ⁡          α                    |        ⋅          sin      2        ⁡    θ    =          |                        5                      cos            2                    ⁡          α          −          9                                      cos            2                    ⁡          α                    ⋅                                    tan            2                    ⁡          θ                          1          +                      tan            2                    ⁡          θ                    |      Now,                               tan          2                ⁡        θ                    1        +                  tan          2                ⁡        θ              =                  4                  cot          2                ⁡        α                    9        +        4                  cot          2                ⁡        α              =                  4                  cos          2                ⁡        α                    9                  sin          2                ⁡        α        +        4                  cos          2                ⁡        α                  =                  4                  cos          2                ⁡        α                    9        −        5                  cos          2                ⁡        α            That leads to product being 4.

kokoszzm · Accepted Answer

Step 1Well I have another solution which is very helpful for exams based on mcq pattern, like the one we give it in india. I have learnt and derived a property in conic sections which says that in a ellipse, the product of perpendiculars from both foci of ellipse to tangent is equal to       b    2              x      2              a      2        +            y      2              b      2        =  1it&#039;s tangent equation is            x      x      1              a      2        +            y      y      1              b      2        =  1 (x1 and y1 are points on ellipse, which is taken   x  1  =  a  cos  ⁡  ϕ and   y  1  =  b  sin  ⁡  ϕ )now tangents is revolving will all possible slopes around ellipse...now to ease our work assume slope =o for a tangent...   α in your equation will be       π    2   for zero slope with y intercept +2 or -2 .put that in your equation and find the product of perpendiculars, as line is horizontal ,perpendicular will be of of equal length. you will find it to be 4. Now as I told product of perpendiculars is       b    2   so   b  =  2 ,now use that general tangent equation and given equation to find value of   a  =  3. You will see that it satisfies the equation and a ellipse. It means we were assuming it right. The parameter in your equation was making the arbitrary line a tangent to ellipse.

Suppose, 2 x cos ⁡ α − 3

Answered question

Answer & Explanation

New Questions in High school geometry