Find the ratio of the area of the region bounded by the parabola and the line segment $PQ$ to the area of...
For the parabola $y=-x^2$,let $a<0,b>0,$ $P(a,-a^2),Q(b,-b^2)$.Let $M$ be the mid point of $PQ$ and $R$ be the point of intersection of the vertical line through $M,$ with the parabola.Find the ratio of the area of the region bounded by the parabola and the line segment $PQ$ to the area of triangle $PQR$
By integration,i found the area of the region bounded by the parabola and the line segment $PQ=frac{-1}{3}(a^3-b^3)$ and the point $R=(frac{a+b}{2},-(frac{a+b}{2})^2)$
I am stuck here.
conic-sections
asked Nov 26 at 13:16
user984325
19112
add a comment |
For the parabola $y=-x^2$,let $a<0,b>0,$ $P(a,-a^2),Q(b,-b^2)$.Let $M$ be the mid point of $PQ$ and $R$ be the point of intersection of the vertical line through $M,$ with the parabola.Find the ratio of the area of the region bounded by the parabola and the line segment $PQ$ to the area of triangle $PQR$
By integration,i found the area of the region bounded by the parabola and the line segment $PQ=frac{-1}{3}(a^3-b^3)$ and the point $R=(frac{a+b}{2},-(frac{a+b}{2})^2)$
I am stuck here.
conic-sections
asked Nov 26 at 13:16
user984325
19112
add a comment |
For the parabola $y=-x^2$,let $a<0,b>0,$ $P(a,-a^2),Q(b,-b^2)$.Let $M$ be the mid point of $PQ$ and $R$ be the point of intersection of the vertical line through $M,$ with the parabola.Find the ratio of the area of the region bounded by the parabola and the line segment $PQ$ to the area of triangle $PQR$
By integration,i found the area of the region bounded by the parabola and the line segment $PQ=frac{-1}{3}(a^3-b^3)$ and the point $R=(frac{a+b}{2},-(frac{a+b}{2})^2)$
I am stuck here.
conic-sections
asked Nov 26 at 13:16
user984325
19112
For the parabola $y=-x^2$,let $a<0,b>0,$ $P(a,-a^2),Q(b,-b^2)$.Let $M$ be the mid point of $PQ$ and $R$ be the point of intersection of the vertical line through $M,$ with the parabola.Find the ratio of the area of the region bounded by the parabola and the line segment $PQ$ to the area of triangle $PQR$
By integration,i found the area of the region bounded by the parabola and the line segment $PQ=frac{-1}{3}(a^3-b^3)$ and the point $R=(frac{a+b}{2},-(frac{a+b}{2})^2)$
I am stuck here.
conic-sections
conic-sections
asked Nov 26 at 13:16
user984325
19112
asked Nov 26 at 13:16
user984325
19112
asked Nov 26 at 13:16
user984325
19112
asked Nov 26 at 13:16
user984325
19112
asked Nov 26 at 13:16
user984325
19112
19112
add a comment |
add a comment |
1 Answer
1
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If $H$ is the projection of $P$ on line $RM$ and $K$ is the projection of $Q$ on line $RM$, then the area of triangle $PQR$ is given by
$$
A_{PQR}={1over2}RMcdot PH+{1over2}RMcdot QK
={1over2}RMcdot (b-a).
$$
Hence you only need to compute
$displaystyle RM={a^2+b^2over2}-left({a+bover2}right)^2$.
answered Nov 26 at 17:06
Aretino
22.6k21442
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $H$ is the projection of $P$ on line $RM$ and $K$ is the projection of $Q$ on line $RM$, then the area of triangle $PQR$ is given by
$$
A_{PQR}={1over2}RMcdot PH+{1over2}RMcdot QK
={1over2}RMcdot (b-a).
$$
Hence you only need to compute
$displaystyle RM={a^2+b^2over2}-left({a+bover2}right)^2$.
answered Nov 26 at 17:06
Aretino
22.6k21442
add a comment |
If $H$ is the projection of $P$ on line $RM$ and $K$ is the projection of $Q$ on line $RM$, then the area of triangle $PQR$ is given by
$$
A_{PQR}={1over2}RMcdot PH+{1over2}RMcdot QK
={1over2}RMcdot (b-a).
$$
Hence you only need to compute
$displaystyle RM={a^2+b^2over2}-left({a+bover2}right)^2$.
answered Nov 26 at 17:06
Aretino
22.6k21442
add a comment |
If $H$ is the projection of $P$ on line $RM$ and $K$ is the projection of $Q$ on line $RM$, then the area of triangle $PQR$ is given by
$$
A_{PQR}={1over2}RMcdot PH+{1over2}RMcdot QK
={1over2}RMcdot (b-a).
$$
Hence you only need to compute
$displaystyle RM={a^2+b^2over2}-left({a+bover2}right)^2$.
answered Nov 26 at 17:06
Aretino
22.6k21442
If $H$ is the projection of $P$ on line $RM$ and $K$ is the projection of $Q$ on line $RM$, then the area of triangle $PQR$ is given by
$$
A_{PQR}={1over2}RMcdot PH+{1over2}RMcdot QK
={1over2}RMcdot (b-a).
$$
Hence you only need to compute
$displaystyle RM={a^2+b^2over2}-left({a+bover2}right)^2$.
answered Nov 26 at 17:06
Aretino
22.6k21442
answered Nov 26 at 17:06
Aretino
22.6k21442
answered Nov 26 at 17:06
Aretino
22.6k21442
answered Nov 26 at 17:06
Aretino
22.6k21442
22.6k21442
add a comment |
add a comment |
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